A football player throws a ball with a horizontal distance of 31.0 m so that the other player catches it at the exact height it was initially thrown. If the ball rises and falls by 1.8 m in its trajectory, what is the initial velocity of the ball? Please give the speed and angle!

Answer 1

how much initial speed up, Vi, to go up 1.8 m?

m g h = (1/2)m Vi^2

Vi^2 = 2 g h
Vi^2 = 2 (9.81)(1.8)
Vi = 5.94 m/s vertical component

Now work on horizontal problem
average speed up =5.94/2 = 2.97 m/s
so time rising up = 1.8/2.97 =.606 s
it goes distance 31/2 = 15.5 m while rising
so u, the constant horizontal component of velocity = 15.5/.06 = 25.6 m/s

so in the end tinital velocity is u = 25.6 and v = 5.94
speed = sqrt(u^2+Vi^2) = 26.3 m/s
tan elevation angle = 5.94/25.6
so angle = 13.1 degrees up from horizontal