Question

An object is thrown with an initial speed of 5 m s^-1 with an angle of projection of 30°. What is the height and range reached by the particle?

Answer 1

Given,

Initial speed (u) = 5 ms^-1

The angle of projection θ = 30°

Solution: 

Max height reached (H_max) = `(u^2 sin^2theta)/(2g) = (25 xx (1/2)^2)/(2 xx 9.8) = 0.318` m

Range (R) = `(u^2sin2theta)/g = (25 xx sin 60^circ)/9.8` = 2.21 m