Question

A particle projected with a speed of 50 m/s has a range that is thrice its greatest height. Find the range

Answer 1

Answer:

R = 240 m

Explanation:

Given,

R = 3H ; u = 50m/s

W.K.T, R = u^2sin2ø/g

H = u^2 sin^2ø/2g

R = 3H

u^2(2sinøcosø)/g = 3u^2sin^2ø/2g

2cosø = 3sinø/2

4/3 = tanø

tanø = 4/3 = OP/AD, HYP = root 4^2 + 3^2 = 5

sinø = 4/5

cosø = 3/5

R = u^2(2sinøcosø)/g

R = 50(50)(2)(4/5)(3/5)/10

R = 100(2)(4)(3)/10

R = 240 m