Answer:
[tex]v_0 = 16.82\ m/s[/tex]
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
[tex]s = ut +\dfrac{1}{2}gt^2[/tex]
[tex]0 =v_0 sin 60^0 t - 4.9 t^2[/tex]
[tex]v_o sin 60^0 = 4.9 t[/tex]
[tex]t = \dfrac{v_0 sin 60^0}{4.9}[/tex]
putting the value of t in equation (1)
[tex]25 = v_0 cos 60^0\times \dfrac{v_0 sin 60^0}{4.9}[/tex]
[tex]25 =\dfrac{v_0^2cos 60^0 sin 60^0}{4.9}[/tex]v
[tex]v_0^2 = 282.90[/tex]
[tex]v_0 = 16.82\ m/s[/tex]
Hence, the initial speed of the rock is equal to 16.82 m/s
