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.
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.
To determine the velocity with which the ball left Latisha's hand, we can use the equation of motion for vertical motion:
Vf^2 = Vi^2 + 2aΔy
Where:
Vf = final velocity (0 m/s at the highest point since the ball momentarily stops before falling down)
Vi = initial velocity (what we want to find)
a = acceleration due to gravity (-9.8 m/s^2, assuming no air resistance)
Δy = change in height (7.0 m)
Using this equation, we can rearrange it to solve for Vi:
Vi = √(Vf^2 - 2aΔy)
Since the ball momentarily stops at the highest point, its final velocity Vf is 0 m/s. Now we can substitute these values into the equation:
Vi = √(0 - 2(-9.8)(7.0))
Simplifying further:
Vi = √(0 + 137.2)
Vi = √137.2
Vi ≈ 11.71 m/s
Therefore, the velocity with which the ball left Latisha's hand is approximately 11.71 m/s.