I got the other ones and everyone helped a lot, I'm very grateful...anyway, last one of the day..
A ball is thrown straight up and returns to the thrower's hand after 3s in the air. A second ball if thrown at an agle of 37 degrees with the horizontal. The acceleration of gravity is 9.8 m/s^2.
At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically? Answer in units of m/s.
2007-09-23 05:27:53 · 1 answers · asked by MD12 2 in Science & Mathematics ➔ Physics
The first ball reached its height in 1.5s. Therefore the initial velocity is 1.5*9.8 = 14.7 (m/s).
For the second ball to reach the same height, the vertical component of its velocity must be 14.7 m/s as well. That is to say:
v*sin(37 degree) = 14.7
So v ≈ 24 (m/s)
2007-09-23 18:19:19 · answer #1 · answered by Hahaha 7 · 0⤊ 0⤋