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mina [271]
2 years ago
8

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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iren2701 [21]

Answer:

Option D

The air pressure inside the car is greater than the pressure outside.

Explanation:

When considering airflow over and around a surface, from Bernoulli's equation, air flow regions with higher velocity have a lower pressure, and regions with lower velocity have a higher pressure.

The air outside the convertible is moving faster than the air inside the convertible. This leads to a higher pressure zone just below the surface of the roof (inside the car) causing the roof of the convertible to bulge upwards

4 0
2 years ago
A particularly scary roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radiu
Pepsi [2]

Answer:

v = 10.89\ m/s

Explanation:

given,                          

radius of loop = 12.1 m                              

to find the minimum speed transverse by the rider to not to fall out upside down                                                                

centripetal force = \dfrac{mv^2}{r}

gravitational force  = m g

computing both the equation]

mg = \dfrac{mv^2}{r}

v = \sqrt{rg}

v = \sqrt{12.1 \times 9.8}

v = \sqrt{118.58}

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5 0
2 years ago
A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-g
mezya [45]

Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

Wavelength of red = 656 nm

Wavelength of blue = 486 nm

Angle = 37°

Suppose we need to find the angle between the red and blue light as it leaves the prism

n_{r}=1.572

n_{b}=1.587

We need to calculate the angle for red wavelength

Using Snell's law,

n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}

Put the value into the formula

1.572\sin37=1\times\sin\theta_{r}

\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})

\theta_{r}=71.0^{\circ}

We need to calculate the angle for blue wavelength

Using Snell's law,

n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}

Put the value into the formula

1.587\sin37=1\times\sin\theta_{b}

\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})

\theta_{b}=72.7^{\circ}

We need to calculate the angle between the red and blue light

Using formula of angle

\Delta \theta=\theta_{b}-\theta_{r}

Put the value into the formula

\Delta \theta=72.7-71.0

\Delta \theta=1.7^{\circ}

Hence, The angle between the red and blue light is 1.7°.

8 0
2 years ago
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gladu [14]

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Pressure equation: P = rho g h

h = P/(rho g)

h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)

h = 0.98 m

0.98m is the maximum depth he could have been.

8 0
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Answer:

700J

Explanation:

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