(a) 9.8 m/s^2, downward
There is only one force acting on the ball while it is in flight: the force of gravity, which is
F = mg
where
m is the mass of the ball
g is the gravitational acceleration
According to Newton's second law, the force acting on the ball is equal to the product between the mass of the ball and its acceleration, so
F = mg = ma
which means
a = g
So, the acceleration of the ball during the whole flight is equal to the acceleration of gravity:
g = -9.8 m/s^2
where the negative sign means the direction is downward.
(b) v = 0
Any object thrown upward reaches its maximum height when its velocity is zero:
v = 0
In fact, at that moment, the object's velocity is turning from upward to downward: that means that at that instant, the velocity must be zero.
(c) 8.72 m/s, upward
The initial velocity of the ball can be found by using the equation:
v = u + at
Where
v = 0 is the velocity at the maximum height
u is the initial velocity
a = g = -9.8 m/s^2 is the acceleration
t is the time at which the ball reaches the maximum height: this is half of the time it takes for the ball to reach again the starting point of the motion, so
[tex]t=\frac{1.77 s}{2}=0.89 s[/tex]
So we can now solve the equation for u, and we find:
[tex]u=v-at=0-(-9.8 m/s^2)(0.89 s)=8.72 m/s[/tex]
(d) 3.88 m
The maximum height reached by the ball can be found by using the equation:
[tex]v^2 - u^2 = 2ad[/tex]
where
v = 0 is the velocity at the maximum height
u = 8.72 m/s is the initial velocity
a = g = -9.8 m/s^2 is the gravitational acceleration
d is the maximum height reached
Solving the equation for d, we find
[tex]d=\frac{v^2-u^2}{2a}=\frac{0^2-(8.72 m/s)^2}{2(-9.8 m/s^2)}=3.88 m[/tex]
