Asked by Drew
A tennis ball is thrown horizontally from an elevation of 20.60 m above the ground with a speed of 19.0 m/s.
(a) Where is the ball after 2.00 s for the horizontal and vertical position?
(b) If the ball is still in the air, how long before it hits the ground?
If the ball is still in the air, where will it be with respect to the starting point once it lands?
(a) Where is the ball after 2.00 s for the horizontal and vertical position?
(b) If the ball is still in the air, how long before it hits the ground?
If the ball is still in the air, where will it be with respect to the starting point once it lands?
Answers
Answered by
Scott
a) vertical __ H = (-.5 * g * 2.00^2) + 20.60 __ g is gravitational acceleration
horizontal __ d = 19.0 * 2.00
b) H = 0 when the ball hits the ground
0 = (-.5 * g * t^2) + 20.60
solve for t to find the flight time
horizontal __ d = 19.0 * 2.00
b) H = 0 when the ball hits the ground
0 = (-.5 * g * t^2) + 20.60
solve for t to find the flight time
Answered by
Jerry
With your equation, my vertical and horizontal answers are correct, however the equation for T is wrong. Any other suggestions?
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