You stand at the edge of a cliff, holding a ball out over the edge. You give the ball an initial velocity of 7.00 m/s straight up, releasing the ball when it is 17.0 m above ground level (the level of the ground at the base of the cliff). Use g = 10.0 m/s2, and neglect air resistance.
To find the time it takes for the ball to reach its maximum height, we can use the equation:
v = u + gt
where:
- v is the final velocity (0 m/s at the maximum height),
- u is the initial velocity (7.00 m/s),
- g is the acceleration due to gravity (-10.0 m/s^2), and
- t is the time.
Rearranging the equation to solve for t, we get:
t = (v - u) / g
Substituting the given values:
t = (0 - 7.00) / -10.0
t = 0.70 s
Now, to find the maximum height reached, we can use the equation:
s = ut + (1/2)gt^2
where:
- s is the displacement (maximum height above ground level),
- u is the initial velocity (7.00 m/s),
- t is the time (0.70 s), and
- g is the acceleration due to gravity (-10.0 m/s^2).
Substituting the given values:
s = (7.00)(0.70) + (1/2)(-10.0)(0.70)^2
s = 4.90 - 1.225
s = 3.675 m
Therefore, the maximum height reached by the ball is 3.675 m above ground level.