a tennis ball is thrown vertically upward from the ground and the student gazing out of the window sees it moving upward pass him at 5 m/sec. the window is 10 m above the ground. how high does the ball go above the ground
V^2 = Vo^2 + 2g*d = 5^2 = 25
Vo^2 - 19.6*10 = 25
Vo^2 = 25 + 196 = 221
Vo = 14.87 m/s.
d = (V^2-Vo^2)/2g.
d = (0-221)/-19.6 = 11.28 m.
To determine how high the tennis ball goes above the ground, we can use the equation of motion for an object in free fall.
Let's break down the information given in the problem:
Initial velocity (vi) = +5 m/s (positive because the ball is moving upward)
Final velocity (vf) = 0 m/s (at the highest point, the ball momentarily stops)
Acceleration (a) = -9.8 m/s^2 (negative because the acceleration due to gravity is directed downward)
Initial position (si) = 0 m (the ground level)
Final position (sf) = ? (the height above the ground that we are trying to find)
The equation we will use is:
vf^2 = vi^2 + 2a*(sf - si)
Plugging in the values:
0^2 = (5 m/s)^2 + 2*(-9.8 m/s^2)*(sf - 0 m)
Simplifying:
0 = 25 m^2/s^2 - 19.6 m/s^2 * sf
Rearranging the equation to solve for sf:
sf = 25 m^2/s^2 / 19.6 m/s^2
sf ≈ 1.28 meters
Therefore, the ball goes approximately 1.28 meters above the ground.