Question

You toss a ball back to a group of people playing on a raised court it you were standing 2 m horizontally away from the court, and tossed the ball with an initial velocity of 5 mis with a 45 angle a.) Find how long it takes the ball to travel to the edge of the court where in lands b.) How high up is the court?

Answer 1

$Time of flight, t = \frac{2 V_{0} sin \theta}{g}$

$Maximum Height reached, H = \frac{V_{o}^{2} sin^{2} \theta}{2 g}$

Where,
V_o is the initial Velocity,
sin θ is the component along y-axis,

a) Time of flight, t = 2 v₀ sin$\Theta$/g

= 2 x 5 x sin 45^o/9.8

= 0.721 sec

b) Maximum Height reached, H

= V_o² sin$\Theta$ /2g

= 5² sin²45 /2 x 9.8

= 0.638 m height