Question

A baseball is popped straight up into the air and has a hang-time of 6.25s. Determine the height to which the ball rises before it reaches its peak.(Hint : the time to rise to the peak is one half the total hang time )

Answer 1

Answer:

Explanation:

At the maximum height of the ball, the velocity will be zero. This will also take half the total time, 3.125 s. 

Now use the equation... 

Δd = vf*t - 1/2at^2 

By setting the final time at the moment of the apex of the height, vf becomes 0. 

So the equation will become... 

Δd = - 1/2at^2 

The equation now resembles the change in height from the ground to the maximum height 

Δd = - 1/2at^2 = - 1/2 (-9.8 m/s^2) (3.125 s)^2 = 47.8 m

Answer 2

48 Meters

Given:

Acceleration = - 9.8 meter / sec^2

Final Velocity = 0 m/sec

t = 3.13 seconds

To Find:

d = ?

Solving:

First using:

Final Velocity = Initial Velocity + acceleration * time

Now substituting the values in the formula:

0 meter/sec = Vi + (-9.8 meter/ second^2) * (3.13 seconds)

0 m/sec = Vi - 30.674 m/sec

Vi = 30.674 m/sec

Rounding we get = 30.7 m/sec

Now using the formula:

vf^2 = vi^2 + 2*a*d

Substituting the values in this formula:

(0 m/sec)^2 = (30.7 m/sec)^2  + 2*(-9.8 m/s^2)*(d)

0 m^2/sec^2 = (940 m^2/sec^2) + (-19.6 m/sec^2)*(d )

-940 m^2/s^2 = (-19.6 m/s^2) * d

d = (-940 m^2/sec^2)/(-19.6 m/sec^2)

d = 48.0 meters

Therefore, the distance traveled is 48 meters.