1.

A projectile has a range of 50m and Reaches a maximum of 10m. what is the elevation of the projectile?

Answer»

We know that Horizontal range R = u^2sin 2theta/g --------------- (1)

We know that Maximum height H = u^2sin^2 theta/2g ------------- (2)

On solving (1) and(2) we get

R/H = 4/tan theta --------------------- (3)

Given, R = 50m, Height = 10, substitute in (3), we get

50/10 = 4/tan theta

tan theta =4/5

theta = 38.6598 degrees.



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