Question

A stone thrown off a bridge 10m above a river has an initial velocity of 8ms at an angle of 25 degrees above the horizontal. What is the range of the stone?

Answer 1

Explanation:

It is given that,

Height above which the stone was thrown, h = 10 m

Initial velocity of the stone, u = 8 m/s

Angle above the horizontal, $\theta=25^{\circ}$

The horizontal component of velocity is, $u_x=v\ cos\theta=8\ cos(25)=7.25\ m/s$

The vertical component of velocity is, $u_y=v\ sin\theta=8\ cos(25)=3.38\ m/s$

Let t is the time of flight in vertical motion. The second equation of motion is :

$h=u_yt-\dfrac{1}{2}gt^2$

$10=3.38t-\dfrac{1}{2}\times 9.8t^2$      

t = 0.34 seconds

Let s is the range of the stone. It can be calculated as :

$s=u_x\ cos\theta \times t$

$s=7.25\times 0.34$

s = 2.46 meters

So, the range of the stone is 2.46 meters. Hence, this is the required solution.