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Grade 11Mechanics

The velocity of the projectile when it is at the greatest height is (2)^1/2÷(5)^1/2 times its velocity when it is at half of its greatest height. Determine its angle of projection.

Profile image of Vanshika
8 Years agoGrade 11
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1 Answer

Profile image of Arun
8 Years ago
The horizontal speed is the same at any time in the trajectory. 
u = total speed at 1/2 max height 
v = horizontal speed = u*sqrt(2/5) 

w = vertical speed at 1/2 max height. Use c^2 = a^2 + b^2 to find an expression for w. 
u^2 = u^2*(2/5) + w^2 
w^2 = u^2*(1 - 2/5) 
w = u*sqrt(3/5) 

We now need to find the initial vertical speed. 
H = maximum height 
H/2 = (1/2)*g*t^2 
H = (1/2)*g*T^2 
T/t = sqrt(2) 

w = g*t 
W = g*T .... W = w*T/t = w*sqrt(2) = u*sqrt(6/5)

tan (Theta) = W/v = sqrt(6/5)/sqrt(2/5) = sqrt(3) 

So theta is 60 degrees