Question

A ball is thrown into the air with an upward velocity of 50.0 m/s at t=0s. If acceleration due to gravity is 9.8m/s2, FIND:
The velocity of the ball at:
t = 4.0s
t = 6.0s
how long it takes to reach its maximum height
the maximum height reached by the ball
the acceleration of the ball at its maximum height
The displacement of the ball during the second second of its flight.

Answer 1

$$v(4)=50-9.8(4)=10.8\frac{m}{s}$$

$$v(6)=50-9.8(6)=-8.8\frac{m}{s}$$

$$t=\frac{50}{9.8}=5.1\ s$$

The maximum height reached by the ball 

$$H=\frac{50^2}{2\cdot 9.8}=128\ m$$

The acceleration of the ball at its maximum height 

$$a=g=-9.8\frac{m}{s^2}$$

The displacement of the ball during the second second of its flight

$$h(2)-h(1)=(50(2)-0.5(9.8)(2)^2)-(50(1)-0.5(9.8)(1)^2)$$ $$h(2)-h(1)=35.3\ m$$