Question

A baseball initially at a height of 5 feet is hit at an angle of 30 degrees to the horizontal at an initial velocity of 150 feet/sec. Will the baseball clear a fence at a height of 15 feet at a distance of 400 feet?

Answer 1

Answer:

Yes

Explanation:

The vertical and horizontal velocity components of the baseball is

$v_v = 150sin(30^0) = 150*0.5 = 75 ft/s$

$v_h = 150cos(30^0) = 150*0.866 = 129.9$

Assume no air resistance, there's no force that affects the ball horizontally. Its horizontal velocity remain constant. So if the wall is 400 feet far from the hit point, the time it takes to travel 400 ft at 129.9 ft/s speed is

$t = s_h/v_h = 400 / 129.9 = 3.08 s$

This is also the time it travels vertically, which is affected by gravity at 9.81 m/s2. So the distance it travels in vertical direction is

$s_v = s_0 + v_vt - gt^2/2$

$s_v = 5 + 75*3.08 - 9.81*3.08^2/2=189.4$

Since 189.4 ft > 15 ft. The ball will be able to clear the fence at 15 ft high and distance of 400 ft.