Question

A rock is thrown at an angle of 60∘ to the ground. If the rock lands 25m away, what was the initial speed of the rock? (Assume air resistance is negligible. Your answer should contain the gravitational constant ????.)

Answer 1

Answer:

$v_0 = 16.82\ m/s$

Explanation:

given,

angle at which rock is thrown = 60°

rock lands at distance,d = 25 m

initial speed of rock, = ?

In horizontal direction

distance = speed x time

d = v₀ cos 60° t

25 = v₀ cos 60° t............(1)

now,

in vertical direction

displacement in vertical direction is zero

using equation of motion

$s = ut +\dfrac{1}{2}gt^2$

$0 =v_0 sin 60^0 t - 4.9 t^2$

$v_o sin 60^0 = 4.9 t$

$t = \dfrac{v_0 sin 60^0}{4.9}$

putting the value of t in equation (1)

$25 = v_0 cos 60^0\times \dfrac{v_0 sin 60^0}{4.9}$

$25 =\dfrac{v_0^2cos 60^0 sin 60^0}{4.9}$v

$v_0^2 = 282.90$

$v_0 = 16.82\ m/s$

Hence, the initial speed of the rock is equal to 16.82 m/s