Question

An arrow is thrown in the air. It's time of flight is 5 s and the range is 200 m. Determine (I) the vertical component of the velocity of projection (ii) angle made with horizontal.(iii) maximum height(IV) horizontal component.

Answer 1

To determine horizontal component of velocity 

as we know 

R = u²sin2θ/g -----1 

T = 2usinθ/g -----2 

now expanding the equation for range R , we get R = 2u²sinθcosθ/g ----3 

dividing equation 3 by equation 2 

we get [2u²sinθcosθ/g]/[ 2usinθ/g] = R/T 

thus we have 

ucosθ = R/T 

or ucosθ = 200/5 = 40m/s 

as we know ucosθ is the horizontal component of velocity when θ is the angle with the horizontal. 

Comments

Angle on the horizontal?????