Question

A man hits a baseball when it is 4 ft above the ground with an initial velocity of 120 ft/sec. The ball leaves the bat at a 30 degree angle with the horizontal and heads towards the 30 ft fence 350 ft away from home plate. The ball will not clear the fence. True or False.
Please help

Answer 1

et us assume that the origin is the floor right below the 30 ft. fence

To work this one out, we'll start with acceleration and integrate our way up to position.

At the time that the player hits the ball, the only force in action is gravity where: a = g (vector)
ax = 0
ay = -g (let's assume that g = 32.8 ft/s^2. If you use a different value for gravity, change the numbers.

To get the velocity of the ball, we integrate the acceleration
vx = v0x = v0cos30 = 103.92
vy = -gt + v0y = -32.8t + v0sin40 = -32.8t + 60

To get the positioning, we integrate the speed.
x = v0cos30t + x0 = 103.92t - 350
y = 1/2*(-32.8)t² + v0sin30t + y0 = -16.4t² + 60t + 4

If the ball clears the fence, it means x = 0, y > 30

x = 0 -> 103.92 t - 350 = 0 -> t = 3.36 seconds

for t = 3.36s,
y = -16.4(3.36)^2 + 60*(3.36) + 4
= 20.45 ft

which is less than 30ft, so it means that the ball will NOT clear the fence.


Just for fun, let's check what the speed should have been :)

x = v0cos30t + x0 = v0cos30t - 350
y = 1/2*(-32.8)t² + v0sin30t + y0 = -16.4t² + v0sin30t + 4

x = 0 -> v0t = 350/cos30
y = 30 ->
-16.4t^2 + v0t(sin30) + 4 = 30
-16.4t^2 + 350sin30/cos30 = 26
t^2 = (26 - 350tan30)/-16.4
t = 3.2s

v0t = 350/cos30 -> v0 = 350/tcos30 = 123.34 ft/s

So he needed to hit the ball at at least 123.34 ft/s to clear the fence.

You're welcome, Thanks please :)