Q.

If time of flight of a projectile is 10 s. Range is 500 m. The maximum height attained by it will be

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a

125 m

b

50 m

c

100 m

d

150 m

answer is A.

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Detailed Solution

Time of flight, T=2usinθg=10 s⇒ usinθ = 50 ms-1∴  H=u2sin2θ2g=(usinθ)22g=50×502×10=125 m
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If time of flight of a projectile is 10 s. Range is 500 m. The maximum height attained by it will be