Question

A ball is thrown with an initial velocity of 20 m/s at an angle of 60° above the horizontal. If we can neglect air resistance, (a) what is the horizontal component of its instantaneous velocity at the exact top of its trajectory? (b) how long is the ball in flight before it hits the ground?(c) how far has the ball traveled in the horizontal direction when it lands?

Answer 1

Answer:

10 m/s

1.87914 s

18.7914 m

Explanation:

v = Initial velocity = 20 m/s

$\theta$ = Angle = 60°

Horizontal component is given by

$v_x=20cos60\\\Rightarrow v_x=10\ m/s$

The horizontal component is 10 m/s

y direction final displacement is zero

$s=vsin \theta t+\frac{1}{2}at^2\\\Rightarrow 0=20sin60+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2\times 20sin 60}{9.81}}\\\Rightarrow t=1.87914\ s$

The time the ball is in the air is 1.87914 s

Range is given is by

$R=vcos\theta t\\\Rightarrow R=10\times 1.87914\\\Rightarrow R=18.7914\ m$

The range is 18.7914 m