Suppose the body is projected from the ground at 24.5 m/s, 60* above the horizontal. Find the (a) time of flight, (b) range, and (c) maximum height. hel (ᗒᗣᗕ)՞

a. t = 4.33 s

b. R = 53.04 m

c. H = 22.97 m

Given

vo=initial velocity=24.5 m/s

Angle = θ = 60°

Required

(a) time of flight, (b) range, and (c) maximum height.

Solution

Parabolic motion

a. time to flight

t = (2vo sin θ) / g

t = (2 x 24.5 x sin 60) / 9.8

t = 4.33 s

b. range (horizontal range) :

R = (vo² sin2θ )/g

R =(24.5² sin 120)/9.8

R = 53.04 m

c. maximum height

H = (vo²sin²θ)/2g

H = (24.5²sin²60)/2.9.8

H = 22.97 m

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