Question 34
Anna Litical drops a ball from rest
from the top of 78.4 m high cliff. How much time will it take for the ball to
reach the ground and at what height will the ball be after each second of the motion?
Reference: Physics Classroom.com
Solution:
Initial velocity of ball =
0 m s-1 (at rest)
Total distance travelled
(fallen) by ball = 78.4 m
Take the acceleration due
to gravity (downward), a = 9.8 m s-2
Let the time at which the
ball reaches the ground be t.
Consider the equation of
uniformly accelerated motion:
s = ut + ½ at2
78.4 = 0 + ½×9.8×t2
Time taken to reach
ground, t = √[78.4 / (0.5×9.8)] = 4.0 s
We also need to find the
height reached after each second (i.e. t = 1, 2, 3, 4 s)
Initial height of the ball
= 78.4 m
The distance fallen by the
ball can be obtained from s = ut + ½ at2 for each second.
At any time t, height of
the ball = 78.4 – Distance fallen by ball at time t
Let the distance fallen by
the ball at time t be s.
s = ut + ½ at2
= 0 + ½×9.8×t2
s = ½ ×9.8 × t2
Using the formula above,
at t = 1 s, distance fallen
= 4.9 m, height of ball = 78.4 – 4.9 = 73.5 m
at t = 2 s, distance fallen
= 19.6 m, height of ball = 78.4 – 19.6 = 58.8 m
at t = 3 s, distance fallen
= 44.1 m, height of ball =78.4 – 44.1 = 34.3 m
at t = 4 s, distance fallen
= 78.4 m, height of ball = 78.4 – 78.4 = 0 (reached ground)
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