Question

"A ball is thrown with an initial speed of 8m/s from a roof top that is 25.0m above ground. The ball is in 'air' for 2.0 seconds and then lands 5.0m in front of the house. 1. Find the horizontal and vertical components of the initial velocity of the ball. 2. Find the maximum height of the ball above ground" - I'm totally missing how to get the horizontal and vertical components - I assumed you need the angle?

Answer 1

Time up and down is 2 s. Range = 5 m. Therefore, since there is no acceleration in the x direction, d = vt, so v = d/t = 5m/2s = 2.5m/s. In the y direction, t = 2s, y initial is 25m, y final = 0. Also, we know v initial in the x-direction is 2.5m/s, = v initial cos(theta) = 8m/s cos(theta). We can solve for theta... theta is cos^-1( 2.5/8) = 71 degrees. Now we can find vy initial and so on.... need more?

Answer 2

Wouldn't that have to be 71º pointing down because it's thrown with a speed of 8m/s, is in air for 2s, and lands 5m away? Wouldn't the maximum height then just equal the height from which it is thrown?

Answer 3

This problem doesn't seem to be too self consistent - if you just drop the ball from 25m height it hits the ground in 2.2s, so it must have an initial velocity down to get there in 2.0s, but if it hits 5m from the house in 2.0s it must have a velocity component of 2.5 m/s in the x-direction. If the initial velocity is 8, that would be the hypotenuse of a right triangle with the x-side being 2.5, giving the y-side as 7.6m/s, which would be vy initial... but that's impossible because it would then reach the ground in 1.6s

Answer 4

idk.