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2 years ago

How do you do projectile motion problems

Physics

2 answers:

skelet666 [1.2K]2 years ago

8 0

Here’s an example

Example 1

CSI discovers a car at the bottom of a 72 m cliff. How fast was the car going if it landed 22m horizontally from the cliff’s edge? (Note that the cliff is flat, i.e. the car came off the cliff horizontally).

v= ? [m/s]

Given: h=Δy=72 m

d=Δx=22 m

g=10.0 m/s2

Equation: h=viyt+12gt2 and d=vixt

Plug n’ Chug:

Step 1: Calculate the time required for the car to freefall from a height of 72 m.

h=viyt+12gt2 but since viy=0, the equation simplifies to h=12gt2 rearranging for the unknown variable, t, yields

t=2hg‾‾‾√=2(72 m)10.0 m/s2‾‾‾‾‾‾‾‾‾‾√=3.79 s

Step 2: Solve for initial velocity:

vix=dt=22 m3.79 s=5.80 m/s

The answer is 5.80 m/s.

PtichkaEL [24]2 years ago

7 0

The key projectile motion is that gravity allows downward only

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The point on the graph that lies on the y-axis (vertical axis) is called the y-intercept. What does the y-intercept tell you abo

jekas [21]

Answer:

The starting position of the runner.

Explanation:

When you look at the graph, you can see that the first point on the graph is twenty on the y-axis.

The runner starts at twenty, and ends at thirty.

Therefore, the runner starts at twenty on the y-axis, so it's the starting position of the runner.

7 0

2 years ago

You and your friend throw balloons filled with water from the roof of a several story apartment house. You simply drop a balloon

Aleks [24]

Answer:

Height = 53.361 m

Explanation:

There are two balloons being thrown down, one with initial speed (u1) = 0 and the other with initial speed (u2) = 43.12

From the given information we make the following summary

$u_{1}$ = 0m/s

$t_{1}$ = t

$u_{2}$ = 43.12m/s

$t_{2}$ = (t-2.2)s

The distance by the first balloon is

$D = u_{1} t_{1} + \frac{1}{2} at_{1}^2$

where

a = 9.8m/s2

Inputting the values

$D = (0)t + \frac{1}{2} (9.8)t^2\\ D = 4.9t^2$

The distance traveled by the second balloon

$D = u_{2} t_{2} + \frac{1}{2} at_{2}^2$

Inputting the values

$D = (43.12)(t-2.2) + \frac{1}{2} (9.8)(t-2.2)^2$

simplifying

$D = 4.9t^2 + 21.56t -71.148$

Substituting D of the first balloon into the D of the second balloon and solving

$4.9t^2 = 4.9t^2 + 21.56t -71.148 \\ 21.56t = 71.148\\ t = 3.3s$

Now we know the value of t. We input this into the equation of the first balloon the to get height of the apartment

$D = 4.9(3.3)^2\\ D = 53.361 m$

7 0

2 years ago

The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vert

zimovet [89]

Answer:

v=5.86 m/s

Explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

$F=\dfrac{mv^2}{r}$

v is the maximum speed

$v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s$

Hence, the maximum speed of the ball is 5.86 m/s.

3 0

2 years ago

The different in size of each of the rope's pullers, correspond to a difference in the magnitude of the applied force, such that

olga55 [171]

Answer:

F = - 50 N

Hence, the magnitude of resultant force is 50 N and its direction is leftwards.

Explanation:

The magnitude of the resultant force is always equal to the sum of all forces. While, the direction of resultant force will be equal to the direction of the force with greater magnitude:

$Resultant\ Force = F = F_{1} - F_{2}$

considering right direction to be positive:

F₁ = Force applied on right rope = 150 N

F₂ = Force applied on left rope = 200 N

Therefore, the resultant force can be found by using these values in equation:

$F = 150\ N - 200\ N$

<u>Hence, the magnitude of resultant force is 50 N and its direction is leftwards.</u>

5 0

2 years ago

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

$F_{No}= 7771.125 \ N$

$F_p = 2.2*10^{6} N$

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

3 years ago