A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $$3$$ $$ms-2$$ for $$0.5$$ min. If the maximum height reached by it is $$80$$ m, then the angle of projection is (g$$ $$=$$ $$10$$ $$ms-2)

Question

A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $$3$$ $$ms-2$$ for $$0.5$$ min. If the maximum height reached by it is $$80$$ m, then the angle of projection is (g$$ $$=$$ $$10$$ $$ms-2)

Infinity Learn · Accepted Answer

$$H=u2sin2$$θ$$2g$$ or $$80=u2sin2$$θ$$2$$×$$10or$$ $$u2sin2$$θ$$=1600$$ or u $$sin$$θ $$=40$$ $$ms-1Horizontal$$ velocity $$=$$ u $$cos$$θ$$=3$$ × $$30=90$$ $$ms-1usin$$θucosθ$$=4090or$$ $$tan$$θ$$=49or$$ θ$$=$$$$tan$$−$$149$$

A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3 {ms}^{- 2}$ for 0.5 min. If the maximum height reached by it is 80 m, then the angle of projection is $(g = 10 {ms}^{- 2})$

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A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of 3 ms-2 for 0.5 min. If the maximum height reached by it is 80 m, then the angle of projection is (g = 10 ms-2)

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A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3 {ms}^{- 2}$ for 0.5 min. If the maximum height reached by it is 80 m, then the angle of projection is $(g = 10 {ms}^{- 2})$