Because of the symmetry in the path of a projectile, it will return to your hand with a speed of 40 m/s downward. Its total displacement is 0 since it returns to starting position.
To find the distance traveled, multiply the height it reaches by 2. Find the height using kinematic equation:
V^2 = V0^2-2gy
where y = h and and V^2 =0 because velocity is 0 at the top of its path.
You throw a baseball straight up at 40 m/s and it reaches its peak after 4.00 seconds. What is its speed when it falls back into your hand? What is its displacement? What distance has it traveled?
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