Projection Under uniform Acceleration

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Question

A body is projected from height of 60 m with a velocity 10 ms^-1 at angle 30⁰ to horizontal. The time of flight of the body is [g =10 ms^-2]

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Solution

$\large h = - u\sin \theta t + \frac{1}{2}g{t^2}$

$\large h = \frac{1}{2}g{t^2} - u\sin \theta t$

$\large 60 = 5{t^2} - 10\frac{1}{2}t$

$\large 5{t^2} - 5t - 60 = 0 \Rightarrow {t^2} - t - 12 = 0$

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Projection Under uniform Acceleration

Remember concepts with our Masterclasses.

A body is projected from height of 60 m with a velocity 10 ms-1 at angle 300 to horizontal. The time of flight of the body is [g =10 ms-2]

Ready to Test Your Skills?

free study material

A body is projected from height of 60 m with a velocity 10 ms^-1 at angle 30⁰ to horizontal. The time of flight of the body is [g =10 ms^-2]