Answer:
a) While in flight, the acceleration is 9.8 m/s² downward.
b) The velocity of the ball when it reaches its maximum height is 0.
c) The initial velocity of the ball is 9.8 m/s.
d) The maximum height the ball reaches is 4.9 m.
Explanation:
The equations for the height and velocity of the ball are as follows:
y = y0 + v0 · t +1/2 · g · t²
v = v0 + g · t
Where:
y = height at time t
y0 = initial position
v0 = initial velocity
t = time
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)
v = velocity at time t
a) Since the only force acting on the ball is the gravity (neglecting air resistance), while in flight, the acceleration will be 9.8 m/s² downward.
b) The velocity of the ball at maximum height is 0, because, in that moment, the ball does not go up nor down. Immediately after, the ball will start to fall with negative velocity.
c) We can use the equation of height to obtain v0, knowing that at t = 2 s, y = 0.
y = y0 + v0 · t +1/2 · g · t² (y0 = 0 because the origin of the frame of reference is located at the throwing point).
0 = 0 m + v0 · 2.00 s - 1/2 · 9.80 m/s² · (2.00 s)²
0 = 2.00 s · v0 - 19. 6 m
19.6 m / 2.00 s = v0
v0 = 9.8 m/s
The initial velocity is 9.8 m/s
d) Let´s find at which time the ball reaches the maximum speed, using the equation of velocity and the fact that at max-height the velocity is 0:
v = v0 + g · t
0 = 9.8 m/s - 9.8 m/s² · t
t = 1 s
The height at t = 1 s will be:
y = y0 + v0 · t +1/2 · g · t²
y = 9.8 m/s · 1 s - 1/2 · 9.8 m/s² · (1 s)²
y = 4.9 m
The maximum height the ball reaches is 4.9 m
