Ask question

Ask question

All categories

2 years ago

8

A good quarterback can throw a football at 27 m/s (about 60 mph). If we assume that the ball is caught at the same height from w

hich it is thrown, and if we ignore air resistance, what is the maximum range in meters (which is approximately the same as the range in yards) of a pass at this speed? How long is the ball in the air?

1 answer:

bezimeni [28]2 years ago

8 0

Answer:

The ball was in air for 3.896 s

Explanation:

given,

g = 9.8 m/s², acceleration due to gravity,

If the launch angle is 45°, the horizontal range will be maximum.

The horizontal and vertical launch velocities are equal, and each is equal to

v_h = v cos θ

v_h = 27 × cos 45°

= 19.09 m/s.

The time to attain maximum height is one half of the time of flight.

v = u + at ∵ v = 0 (max. height)

19.09 - 9.8 t₁ = 0

t₁ = 1.948 s

The time of flight is twice of the maximum height time

2 t₁ = 3.896 s

The horizontal distance traveled is

D = v × t

D = 3.896×19.09

= 74.375 m

The ball was in air for 3.896 s

You might be interested in

Irrigation channels that require regular flow monitoring are often equipped with electromagnetic flowmeters in which the magneti

san4es73 [151]

A) $1.36\cdot 10^{-4}T$

The magnetic field at the center of a coil of N turns is given by

$B=\frac{\mu_0 N I}{2R}$

where

I is the current in the coil

N is the number of turns

R is the radius of the coil

Here we have

I = 6.5 A is the current in the coil

N = 100 is the number of turns

$R=\frac{6.0 m}{2}=3.0 m$ is the radius of the coil

Substituting,

$B=\frac{(4\pi \cdot 10^{-7} H/m)(100)(6.5 A)}{2(3.0 m)}=1.36\cdot 10^{-4}T$

B) The force points north

The direction of the force on a positive ion in water can be found by using the right-hand rule. In fact, we have:

- Index finger: direction of motion of the ion --> towards east

- Middle finger: direction of magnetic field --> downward

- Thumb: direction of the force --> towards north

So, the force points north.

C) $3.26\cdot 10^{-23}N$

The magnitude of the magnetic force on a charged particle moving perpendicularly to the field is

$F=qvB$

where

q is the charge of the particle

v is the velocity

B is the magnitude of the magnetic field

In this case, we have

$q=+e=1.6\cdot 10^{-19} C$ is the charge

$v=1.5 m/s$ is the velocity

$B=1.36\cdot 10^{-4}T$ is the magnetic field strength

Substituting,

$F=(1.6\cdot 10^{-19} C)(1.5 m/s)(1.36\cdot 10^{-4}T)=3.26\cdot 10^{-23}N$

8 0

2 years ago

Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

schepotkina [342]

If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!

5 0

2 years ago

Under the Big Top elephant, Ella (2500 kg), is attracted to Phant, the 3,000 kg

Vladimir [108]

Under the Big Top elephant, Ella (2500 kg), is attracted to Phant, the 3,000 kg elephant. They are separated by 8

4 0

2 years ago

A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mir

Semmy [17]

Answer:

The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Explanation:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{0.24}=\dfrac{1}{0.25}+\dfrac{1}{u}$

$\dfrac{1}{u}=\dfrac{1}{0.24}-\dfrac{1}{0.25}$

$\dfrac{1}{u}=\dfrac{1}{6}$

$u=6\ m$

We need to calculate the magnification

Using formula of magnification

$m=-\dfrac{v}{u}$

Put the value into the formula

$m=-\dfrac{0.24}{-6}$

$m=0.04$

We need to calculate the height of the object

Using formula of magnification

$m=\dfrac{h'}{h}$

$h=\dfrac{0.080}{0.04}$

$h=2\ m$

A convex mirror produce a virtual and upright image behind the mirror.

Hence, The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

6 0

2 years ago

Read 2 more answers

A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a hor

nalin [4]

Answer:

$u_x=38.13\ m/s$

Explanation:

Given that initially ball moves in the horizontal direction ,it means that the velocity in the vertical direction is zero.

Horizontal distance = 13 m

Vertical distance = 57 cm

Lets take time to cover 57 cm distance in vertical direction is t.

We know that g is the constant acceleration in the vertical direction so we can apply the equation of motion in the vertical direction.

$S=u_yt+\dfrac{1}{2}gt^2$

Here $u_y=0$

S= 57 cm

$0.57=0\times t+\dfrac{1}{2}\times 9.81\times t^2$

t=0.34 s

Now in the horizontal direction

$x=u_xt$

Here x=13 m

t= 0.34 s

So

$13=u_x\times 0.34$

$u_x=38.13\ m/s$

So the initial speed of ball is 38.13 m/s.

7 0

2 years ago