Question

A man standing on a bridge throws a stone horizontally with a speed of 20m/s. The stone hits the water below 3 seconds later. How high is the bridge?

Answer 1

 h = 1/2 * g t^2 
= 1/2 * 9.8 * 3^2 
= 44.1 m 

d = vx * t = 20 * 3 = 60 m 
Ans: 44.1 m, 60 m 

Answer 2

Answer:

The bridge is 44.15 m high

Explanation:

This is a case of projectile motion, the stone is projected into the air and is acted upon by force of gravitation until  it ends up in the water.

Projectile motion

This can be described as a motion of an object affected by gravitational force when projected or thrown into the air and it normally has a parabolic pathway. An Example is a  motion of projected missile.

The figure attached shows the man and the projectile motion of the stone.

Calculations

The equation of motion (equation 1) can be used in calculating the height of the bridge;

$S = ut + \frac{1}{2} gt^{2}$........................................................... 1

where S is the distance

          u is the initial velocity

          g is the acceleration due to gravity

          t is the time

The stone was at rest before thrown, so the initial velocity is 0 m/s;

Given u = 0 m/s

          t = 3 secs

          g (constant) = 9.81 $m^{2}$/s

the height can be calculated by substituting the values above into equation 1

S = (0 m/s)(3 secs) + $\frac{1}{2}$(9.81 m/ $s^{2}$)$(3secs)^{2}$

S = 0 + 44.15 m

S = 44.15 m

Therefor the height of the bridge is 44.15 m