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horizontal range = 200m, maximum height= 25m find the angle of projection.


Sagot :

Answer:

θ = 26.6°

Explanation:

Formula for maximum height is;

h = u²sin²θ/2g

We are told maximum height is 25 m, thus;

u²sin²θ/2g = 25 - - - - (eq 1)

Formula for horizontal range is;

R = u²sin2θ/g

In Mathematics, sin2θ = 2sinθcosθ

So; R = 2u²sinθcosθ/g

R = 2u²sinθcosθ/g

We are given Horizontal range = 200 m.

Thus;

2u²sinθcosθ/g = 200 - - - - (eq 2)

Divide eq 1 by eq 2 to get;

200/25 = 4cosθ/sinθ

8 = 4cosθ/sinθ

4/8 = sinθ/cosθ

0.5 = tan θ

θ = tan^(-1) 0.5

θ = 26.6°