A ball is thrown straight upward and returns to the thrower's hand after 2.88 s in the air. A second ball is thrown at an angle of 34.0 degrees with the horizontal. The acceleration of gravity is 9.81 m/s2 .
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.
2007-09-24 11:13:26 · 1 answers · asked by fatiimaq08 3 in Science & Mathematics ➔ Physics
For the first ball, the in-air time is 2.88 s. That means it reaches the maximum height in 1.44s. Thus its initial speed is: 1.44*9.81 = 14.1 (m/s)
Let the second ball thrown with initial velocity V. Its vertical component is then V*sin34.0°. For the second ball to reach the same height as the one thrown vertically, the vertical component must be the same as the one thrown vertically. That is:
V*sin34.0° = 14.1 (m/s)
V = 25.3 (m/s)
2007-09-27 14:08:56 · answer #1 · answered by Hahaha 7 · 0⤊ 0⤋