A stone was projected at an angle of 40o and initial velocity of 20m/s. (a.)Determine the time of flight  (b)Maximum height.

A.) 1.3 seconds

B.) 0.42 m

Explanation:

A.) You are given the angle of projection to be 40 degrees and initial velocity of 20m/s. 

At vertical component

U = Usin 40 that is,

U = 20sin40

Using the first equation of motion under gravity

V = U - gt

Let V = 0

0 = UsinØ - gt

gt = UsinØ

t = UsinØ/g

Where U = 20 m/s

Ø = 40 degree

g = 9.8 m/s^2

Substitutes all the parameters into the formula

t = 20sin40/9.8

t = 1.3 seconds

Total time of flight T = 2t

T = 2 × 1.3 = 2.6 s

B.) To calculate the maximum height,

You will use the formula

V^2 = U^2 - 2gH

At maximum height, V = 0

2gH = Usin^2Ø

H = Usin^2Ø/ 2g

Substitutes all the parameters into the formula

H = 20 sin^2(40) ÷ 2(9.8)

H = 8.2635/19.6

H = 0.42 m

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