The acceleration of gravity is 9.8 m/s 2 . By how much vertical distance does the ball clear the crossbar? Answer in units of m.

Question

To win the game, a place kicker must kick a football from a point 19 m (20.7784 yd) from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 16 m/s at an angle of 59.8 ◦ from the horizontal. The acceleration of gravity is 9.8 m/s 2 . By how much vertical distance does the ball clear the crossbar? Answer in units of m.

Answer 1

First we determine the vertical  and horizontal components of the initial velocity. These are given by

v_0x = (v₀)cosθ  and v_0y = (v₀)sinθ where v₀ = 16 m/s and θ = 59.8°

So v_0x = 16 cos(59.8) = 8.05 m/s

and v_0y = 16 sin(59.8) = 13.83 m/s

The equations of horizontal and vertical motion can be solved independently, so it's a two part problem.

The Horizontal Part (calculating the time it takes the ball to reach the crossbar)

We use v_0x = 8.05 m/s to calculate how long it takes the ball to travel 19 m

Since d = (v_0x)t, this means that

t = d/v_0x = (19 m)/8.05 m/s) = 2.36 s

The Vertical Part (calculating the height of the ball when it reaches the crossbar)

The height of the ball at the time (t) that we just calculated above is given by

h = (v_0y)t – ½gt² = (13.83)(2.36) – ½(9.8)(2.36)² = 32.64 – 27.29 = 5.35 m

Since the crossbar is 3.05 m high, this means the ball clears the crossbar by

5.35 – 3.05 = 2.3 m