Answer:
A. 28.42 m/s
B. 41.21 m.
Explanation:
A. Determination of the initial velocity of the ball:
Time (t) to reach the maximum height = 2.9 s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)
Initial velocity (u) =?
Thus, we can obtain the initial velocity of the ball as follow:
v = u + gt
0 = u + (–9.8 × 2.9)
0 = u – 28.42
Collect like terms
u = 0 + 28.42
u = 28.42 m/s
Therefore, the initial velocity of the ball is 28.42 m/s.
B. Determination of the maximum height reached.
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)
Initial velocity (u) = 28.42 m/s.
Maximum height (h) =?
Thus, we can obtain the maximum height reached by the ball as follow:
v² = u² + 2gh
0² = 28.42² + (2 × –9.8 × h)
0 = 807.6964 + (–19.6h)
0 = 807.6964 – 19.6h
Collect like terms
0 – 807.6964 = – 19.6h
– 807.6964 = – 19.6h
Divide both side by – 19.6
h = –807.6964 / –19.6
h = 41.21 m
Therefore, the maximum height reached by the ball is 41.21 m
