A homr run is hit into the upper deck of a stadium. The baseball lands (it is descending when it lan...?

Question

A homr run is hit into the upper deck of a stadium. The baseball lands (it is descending when it lands) 60ft above the height from which it was hit. The ball left the bat with a speed of 85 mi/hr at an angle of 60 degrees above the horizontal (Ignore air resistance)

a) what is the speed of the ball when it lands?

b) How long was the ball in the air?

c) What is the horizontal distance from home plate (where the ball was hit) to the point where the ball landed? Put your answer to feet.

d) How far (horizontally would it have gone if it had be allowed to fall back to the height it had been hit?

Answer 1

a. First let's find out to what maximum height the ball went using the formula: v^2-u^2=2as

where v=final velocity in m/s
u = the initial velocity in m/s
a= the acceleration of gravity in m/s^2
s= the maximum height reached by the ball in m.

Given: v=0 (because ball stops rising before it starts to fall)
u=85mph or 85*1.601*1000/3600m/s or 37.8 m/s
a=9.8m/s^2

Substitute known values:

0^2-37.8^2=2*(9.8)s
1429=19.6s
s=1429/19.6
=72.9m

If the ball landed 60ft (or 18.3m) above the point where it was
hit, that means that it was 72.9 minus 18.3 or 54.6m below the top of the rise. Let's solve for the vertical component of the speed at that point using the same formula above.

Given: s=54.6m
u=0 (we start from the top of the rise)
a=9.8m/s

Substitute known values:

v^2-0=2*9.8*54.6
v^2=1070
v=32.7m/s

The horizontal component of the speed is:

v=37.8cos60
=18.9m/s

The speed is the the hypotenuse of a right angle with sides 18.9 and 32.7, or the square root of (18.9^2+32.7^2). Pls do the calculation. The angle of the speed from the horizontal is arctan32.7/18.9.

b. the total time can be solved by using the formula:

s=ut+1/2at^2
First solve for the time it takes to fall from the top to the point where it was hit, using the above formula:
Given: s=72.9m
u=0
a=9.8m/s^2
72.9=0+1/2(9.8)t^2

Solve for t .

Then solve for the time it takes to fall from the top to the point where it lands.

s=54.6m
u=0
a=9.8m/s^2

Substitute in the formula s=ut+1/2at^2, and solve for t.

The total time is the sum of the two t's.

c. Use the basic formula s=vt, where s is the horizontal distance, v is the horizontal speed, and t the total time you got in b above.

s=18.9*t

Solve for s since you know the value of t.

d. Solve for s using again the formula s=vt. This time t is the time it takes to reach the top multiplied by 2. Why? Because it takes the same time to reach the top as the time to reach the bottom from the top. Moreover, the time it takes to travel from the point where it was hit to the point where it would have fallen had it been allowed to fall back to the height it had been hit, would be the same as the time it takes to reach the top and fall back to the height where it was hit. Thus to solve for the horizontal distance (or range) multiply the horizontal speed by the total time.

Note: In our calculations, we used the equivalents, viz:

1 mile=1.601 km
1km=1000m
1hr=3600 s
1m=3.28ft

If he answer is required in feet, just use the above figure (1m=3.28ft)

There are some basic things you have to know about projectiles. One of them is that to get the horizontal distance or range, you merely use the horizontal speed and multiply it by the time of travel. No need to worry about acceleration. The speed is constant. The other is what I said about the time it takes to reach the top being equal to the time it takes to reach the bottom.

Finally, it would be a great help if you can draw the trajectory of the projectile (the ball) and indicate distances between points. That I could do easily if you were just sitting beside me.