To solve this problem, we can use the equations of motion for free fall:
a) Let's consider the motion of the ball when it reaches the highest point of its trajectory:
v = u + at
0 = u - 9.8(5)
u = 49 m/s
b) To find out how high the ball rose above its starting point, we can use the equation:
s = ut - (1/2)gt^2
s = (49)(5) - (1/2)(9.8)(5)^2
s = 245 - 122.5
s = 122.5 m
c) To find the velocity of the ball as it passed 5 m above its starting point, we can use the equation:
v^2 = u^2 + 2as
v^2 = (49)^2 + 2(-9.8)(-5)
v^2 = 2401 + 98
v^2 = 2499
v = 49.99 m/s
d) To find the velocity of the ball as it passed 5 m below its starting point, we can use the equation:
v^2 = u^2 + 2as
v^2 = (49)^2 + 2(9.8)(5)
v^2 = 2401 + 98
v^2 = 2499
v = 49.99 m/s
Therefore,
a) The initial velocity of the ball was 49 m/s.
b) The ball rose 122.5 m above its starting point.
c) The magnitude of its velocity as it passed 5 m above the starting point was 49.99 m/s.
d) The magnitude of its velocity as it passed 5 m below the starting point was 49.99 m/s.
A ball is thrown nearly vertically upward from a point near the comice of a tall building. It just misses the cornice on the way down and passes a point 49 m below its starting point 5 sec after it leaves the thrower's hand.
a) What was the initial velocity of the ball?
b) How high did it rise above its starting point?
c) What was the magnitude of its velocity as it passed 5 m above the starting point?
d) As it passed 5 m below the starting point?
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