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Question

A body is projected from the top of a tower of height 32 m with a velocity of 20 m/s at an angle of 37 above the horizontal. The time of flight of the projectile (in seconds) is

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Solution

Time of flight only depends on the vertical component of velocity.
Initial vertical velocity uy=20sin37=20×35=12 m/s (taking upward direction as positive)

Displacement of the body when it reaches the ground h=32 m

Using the second equation of motion,
h=uyt12gt2
32=12t5t2
5t212t32=0
5t220t+8t32=0
5t(t4)+8(t4)=0
(5t+8)(t4)=0
t=4 (since time cannot be negative)
Hence the time of flight is 4 s.

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