A baseball player throws a ball straight up into the air.At a height of 3.00 m above the release point,the ball is moving at 1.25 m/s.What was the ball's initial speed?
A)7.56 m/s
B)7.77 m/s
C)57.2 m/s
D)60.4 m/s

Question

A baseball player throws a ball straight up into the air.At a height of 3.00 m above the release point,the ball is moving at 1.25 m/s.What was the ball's initial speed?
A)​7.56 m/s
B)​7.77 m/s
C)​57.2 m/s
D)​60.4 m/s

Quizplus · Accepted Answer

When the ball is at its maximum height, its velocity is zero. Therefore, the initial velocity of the ball can be found using the following kinematic equation:
v² = u² + 2as
where
v = final velocity = 0 m/s (at the maximum height)
u = initial velocity (what we want to find)
a = acceleration due to gravity = -9.81 m/s² (negative because it acts in the opposite direction to the motion)
s = displacement = 3.00 m (the height reached by the ball)

Substituting the known values, we get
0² = u² + 2(-9.81)(3.00)
Simplifying,
u² = 2(9.81)(3.00)
u² = 58.86
Taking the square root of both sides,
u = 7.677
The initial speed of the ball was 7.77 m/s (to two significant figures). Therefore, the best choice is B.

A Baseball Player Throws a Ball Straight Up into the Air.At