To determine the velocity with which the ball left Latisha's hand, we need to use the kinematic equation for height, which is:
Final velocity squared (v_f^2) = Initial velocity squared (v_i^2) + 2 * acceleration (a) * height (h)
In this case, we assume that the initial height is 0m, so the equation becomes:
v_f^2 = v_i^2 + 2 * a * h
Since the ball is thrown vertically up, the acceleration (a) will be equal to -9.8 m/s^2 (negative because it acts against the motion of the ball). The final height (h) is given as 7.0m.
v_f^2 = v_i^2 + 2 * (-9.8 m/s^2) * 7.0m
Since the ball reaches its highest point at the height of 7.0m, and we're finding the velocity with which it left Latisha's hand (which is the initial velocity), we know that the final velocity (v_f) will be 0 m/s (as it momentarily stops at the highest point).
0^2 = v_i^2 + 2 * (-9.8 m/s^2) * 7.0m
Simplifying the equation:
0 = v_i^2 - 137.2 m^2/s^2
Rearranging the equation:
v_i^2 = 137.2 m^2/s^2
Taking the square root to solve for v_i:
v_i = √137.2 m^2/s^2
v_i ≈ 11.71 m/s
Therefore, the velocity with which the ball left Latisha's hand is approximately 11.71 m/s.
Latisha throws a tennis ball vertically up into the air. It reaches a height of 7.0m. determine the velocity with which the ball left her hand.
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