science
a ball is thrown straight upward and returns to the thrower's hand after 3.00s in the air. A second ball is thrown at an angle of 30.0 deg with the horizontal. At what speed must the second ball be thrown so that it reached the same height as the one thrown vertically?
Reply
Answers
The speed of the second ball can be determined using the equation v^2 = vo^2 + 2gh, where 'v' is the final speed, 'vo' is the initial speed, 'g' is the gravitational acceleration, and 'h' is the height. Since the same height is being reached, h is the same for both cases, so this equation can be solved for vo, which is the initial speed for the second ball. vo = sqrt(2gh), where h is the same height reached by the first ball. Substituting in the values for g (9.8 m/s^2) and h (the height reached by the first ball in 3.00s, which is 4.9 m) yields vo = 11.9 m/s. The initial speed of the second ball must be 11.9 m/s in order to reach the same height as the first ball.
