Anna Litical drops a ball from rest from the top of 78.4 m high cliff.

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)