1) The initial speed is V = sqrt(18^2 + 18^2)
2) µ = arctan(1.0) = 45º
3) The maximum height reached derives from h = V^2(sin^2(µ))/2g.
4) The distance traveled by the ball derives from d = V^2(sin(2µ))/g.
5) From Vf = Vo + gt, the time to reach the maximum height derives from 0 = 18 - 9.8t making t = 1.836sec, telling us that the ball is on the way down at 2.1sec, .264sec from the high point.
The vertical speed at this point is .264(9.8) = 2.58m/s making the net speed sqrt(2.58^2 + 18^2)
I'll leave the height of the ball at 2.1sec. for you to work out now that you have all the data.
A soccer ball is kicked with an initial horizonal velocity of 18 m/s and an initial vertical velocity of 18 m/s.
1) What is the initial speed of the ball?
2)What is the initial angle of the ball with respect to the ground?
3)What is the maximum height the ball goes above the ground?
4) How far from where it was kicked will the ball land?
5)What is the speed of the ball 2.1s after it was kicked? and how high above the ground after it was kicked?
2 answers
A ball is thrown horizontally from a height of 29 m and hits the ground with a speed that is 4 times its initial speed
What is the initial speed
What is the initial speed