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# A ball thrown straight up takes 2.25 s to reach a height of 36.8 m. (a) What was its initial speed? (b) What is its speed at this height? (c) How much higher will the ball go?

Navjyot Kalra
6 years ago
Given:
Height of ball at a particular instant, x = 36.8m .
Time taken by ball to reach height , t = 2.25 s.
Acceleration due to gravity,g = 9.8 m/s2 .
(a) Let us assume that the initial speed of the ball be represented by v0. From equation of kinematics, one can relate the height x with time t and initial speed v0 as:

Since the ball move upwards under the action of gravity, the ball will decelerate such that the deceleration is equal to the magnitude of the acceleration due to gravity, and acts in direction opposite to that with initial velocity.
Therefore, a = -g.
If we assume that at time t = 0 , the ball was thrown upward, then x0 = 0. Substitute the initial conditions and the given values in the equation above to have:

(c) When the ball reaches the maximum height, (say h)its final speed (v) is zero. With same initial speed v0, the maximum height can be evaluated as:
From the equation of kinematics, we have

Where v represents the final speed of the object under consideration, a is its acceleration, v0 the initial speed and t the time for which the object is studied.
From the second equation, the time t is given as:

When the ball reaches the maximum height x = h, v = 0m/s and a = -g
Substitute the given values in the equation above to have
Therefore the height to which the ball rises is 38 m.