Question #179557

A projectile is launched from ground level with the initial velocity of 50.0 m/s . The time of the flight is measured to be 6.00 s. How far from the launch point does the projectile land? Neglect the air resistance and g = 10.0 m/s^2


(A) 120 m

(B) 150 m

(C) 240 m

(D) 300 m


Expert's answer

From the time of the flight find the angle above the horizontal:


t2=vsin⁡θg, θ=arcsin⁡gt2v.\frac t2=\frac{v\sin\theta}{g},\\\space\\ \theta=\arcsin\frac{gt}{2v}.

Find how far the projectile landed:


R=tvcos⁡θ=tvcos⁡[arcsin⁡gt2v]=240 m.R=tv\cos\theta=tv\cos\bigg[\arcsin\frac{gt}{2v}\bigg]=240\text{ m}.


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