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

Step-by-step 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-(1)/(2)gt^2`

`10=3.38t-(1)/(2)* 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 * t`

`s=7.25* 0.34`

s = 2.46 meters

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