Question

A stone is thrown from ground level at an angle of θ above the horizontal such that the ratio of the x-component of its initial velocity to the y-component of its initial velocity is 2. The stone strikes a wall 6 m away and at a height of 2 m above the ground. Calculate the initial speed of the stone

Answer 1

Let us consider the initial velocity of the stone is V, therefore
Horizontal component of the velocity (V_X) = VCos
Vertical component of the velocity (V_Y) = VSin
Now it is given that



Now Horizontal distance covered to reach the wall (R) = 6 m
R = VCos*t
t = R/VCos= 6/VCos26.56 = 6.71/V -----------(1)
Now considering the vertical motion
Height covered by the stone in same time (H) = 2 m
Now using kinematic equation , we can write
H = VSin*t +(1/2)(a)t²
where a is the acceleration = -g = -9.81 m/s²
2 = VSin26.56*(6.71/V) - (1/2)*9.81*(t)²
2 = 3 -4.905*t²
4.905*t² = 1
t = 0.45 s
Now from equation 1
t = 6.71/V
V = 6.71/t = 6.71/0.45
V = 14.861 m/s
Hence the initial velocity of the stone would be 14.861 m/s