A baseball is thrown straight upward with a speed of 30 m/s. (a) How long will it rise? (b) How high will it rise? (c) How long after it leaves the hand will it return to the starting point? (d) When will its speed be 16 m/s?
Answers
Answer:
People often run straight to equations. Provided you make some assumptions (no air resistance, probably not a good approximation, but we really have to make it), you should be able to solve this based on:
Definitions of velocity and acceleration
2. Understanding that the acceleration due to gravity is (approx) 10 m/s^2 DOWN.
At the top, the ball is not going up or down, and so its velocity is instantaneously 0.
Therefore it takes 3 seconds to go up, as that is the time necessary for it to slow down at 10 m/s^2.
The average velocity, when acceleration is constant, is half-way between the beginning and ending velocity. Displacement = average velocity x time = (30/2)*3 = 45 meters.
Then it comes down. The acceleration is still 10 m/s^2. By symmetry, it takes the same time down as it did up. So, it reaches you 6 s after you threw it.
Concept:
We need to apply the kinematic equations given as - v = u+at, v² - u² = 2as.
Given:
Speed of baseball when thrown upwards = 30m/s
Find:
We need to determine the:
(a) time taken by the ball to rise
(b) The height of the ball
(c) Time taken by the ball to return to the starting point
(d) Time taken to have speed 16m/s
Solution:
We know that because the earth is dragging the ball in a downward direction, it will accelerate downward at a rate of -9.8 m/s2 (*but here we will use -10 m/s2) while travelling upward.
final velocity(v) is 0 m/s.]
u(initial velocity)=30 m/s.
v = 0 m/s is the final velocity.
a = g = -10 m/s2 for acceleration.
Therefore, we must determine the height(s) it rises.
Having said that, v² - u² = 2as
s = v² - u²/2a
s = 0 - (30)²/2(-10)
s = 45m is the height of the rise of the ball.
Now, applying v = u+at
0 = 30 + (-10) t
t = 30/10
t = 3secs is the time taken by the ball to rise to the maximum height.
Since the ball accelerates at the same rate or 10 m/s2, it is obvious that it will take the ball 3 seconds to reach the ground from above.
The ball will therefore return to the hand after 2 × 3 = 6 seconds.
If velocity is 16m/s at v = 0
then v = u+at
0 = (16) + (-10)t
t = 16/10
t = 1.6 secs at speed 16m/s
Thus, (a) The time taken by the ball to rise is 3 secs
(b) The height of the ball is 45m
(c) The ball will take 6 seconds to return to the starting point.
(d) The speed of the ball will be 16m/s at 1.6 seconds.
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