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in Physics by (53.6k points)

An object is projected with a velocity of 60m/s in a direction making an angle of 60° with the horizontal. 

Find 1. the maximum height 

2. the time taken to reach maximum height 

3. the horizontal range

1 Answer

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Best answer

u = 60m/s, θ = 60°

1. Maximum height

H = \(\frac {u^2 sin^2 (\theta)}{2g}\)

\(\frac {60^2 sin^2 (60)}{2 \times \,9.8}\)

= 137.74 m.

2. Time to reach maximum height

t = \(\frac {u sin (\theta)}{g}\)

\(\frac {60\times sin (60)}{9.8}\)

= = 5.30s.

3. Range R = \(\frac {u^2 sin (2\theta)}{g}\)

\(\frac {60^2 sin(2 \times 60)}{9.8}\)

= 318.12 m.

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