Question

A ball is thrown horizontally from the top of a tower with speed of12.5 ms_1 strikes the ground 75m away from the tower .lfg=10m ms_2, height of the tower is

Answer 1

Height of tower would be 180m.
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Answer 2

Concept:

We need to apply the third kinematic equation- H = ut + 1/2 gt² and then t was also determined by the distance formula.

Given:

Speed of ball = 12.5 m/s

Distance of the ball when strikes the ground away from the tower is 75m

Acceleration due to gravity = 10 m/s²

Find:

We need to determine the height of the tower.

Solution:

We first need to apply the third kinematic equation of motion represented as H = ut + 1/2 gt² where H is the height of the tower, u is the initial velocity, t is the time and g is the acceleration due to gravity.

We have, u = 0, g = 10

Therefore, the third kinematic equation becomes-

H = 0 + 1/2 (10)t²

H = 5t²

To determine the time, t we need to apply the equation of distance which is represented as-

distance = speed × time

time = distance/ speed

We have, distance, d = 75m and speed = 12.5m/s

Therefore, the equation for a time becomes-

time = 75/12.5

t = 6sec

We have, H = 5t²

Substitute value of t = 6

height of the tower becomes, H = 5(6)²

H = 5 × 36

H = 180m is the height of the tower.

Thus, the height of the tower is 180m.

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