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?

Question

asked Jul 21, 2020 9.7k views

1 Answer

Tsurahman · Answer 1 · 2020-07-26T18:37:40+0000

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.