Question

a body was thrown from top of a tower 60m High above ground at an angle 30° with vertical. Calculate horizontal range. take g=10m/s² ​

Answer 1

Given info : A body was thrown from the top of a tower 60m high above the ground at an angle of 30° with vertical.

To find : The horizontal range of projectile is...

solution : angle made by projectile with vertical is 30° so angle made by projectile with horizontal is (90° - 30°) = 60°. [ see figure ]

let u is the initial velocity of projectile.

vertical component of velocity = usin60°

horizontal component of velocity = ucos60°

using formula,

y = $u_yt+\frac{1}{2}a_yt^2$

⇒60 m = usin60° t + 1/2 × -10 × t²

⇒60 = usin60° t - 5t²....(1)

again, $x=u_xt+\frac{1}{2}a_xt^2$

⇒R = ucos60° t + 1/2 × 0 × t² [ in horizontal, acceleration doesn't exist. ]

⇒t = R/ucos60° ....(2)

from equations (1) and (2) we get,

60 = tan60° - 5R²/u²cos²60° ...[3]

here you see, without velocity we can't find range R. just put the value of u in the above equation and you will get the answer of R.