A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37 above horizontal]. It gets blocked just after reaching a height of 3.0 m. (a) What is the velocity of the football when it first reaches a height of 3.0 m above the level ground? (b) What horizontal distance has the ball travelled when it first reaches a height of 3.0 m above ground?

Refer to the diagram shown below.

Neglect air resistance.

The horizontal component of the launch velocity is

(20 m/s) * cos (37°) = 15.973 m/s

The vertical component of the launch velocity is

(20 m/s) * sin (37°) = 12.036 m/s

The acceleration due to gravity is g = 9.8 m/s².

The time, t s, for the ball to reach a height of 3 m is given by

(12.036 m/s) * (t s) - (1/2) * (9.8 m/s²) * (t s) ² = (3 m)

12.036t - 4.9t² - 3 = 0

2.4543t - t² - 0.6122 = 0

t² - 2.4563t + 0.6122 = 0

Solve with the quadratic formula.

t = (1/2) [2.4563 + / - √ (6.0334 - 2.4488) ]

t = 2.1748 or 0.2815 s

The ball reaches a height of 3 m twice.

The first time it reaches 3 m height is 0.2815 s.

Part a.

The vertical velocity when t = 0.2815 s is

Vy = 12.036 - 9.8*0.2815

= 9.2773 m/s

The horizontal component of velocity is Vx = 15.973 m/s

The resultant velocity is

√ (9.2773² + 15.973²) = 18.47 m/s

Answer:

The velocity at a height of 3.0 m is 18.5 m/s (nearest tenth)

Part b.

The horizontal distance traveled is

d = (15.973 m/s) * (0.2815 s) = 4.4964 m

Answer:

The horizontal distance traveled is 4.5 m (nearest tenth)