A projectile is launched at an angle of 60o with respect to the horizontal from

Question

A projectile is launched at an angle of 60o with respect to the horizontal from the edge of a cliff which is 100 m high. Its initial speed is 30 m/s. The cliff is perfectly vertical and the ground from the cliff outward is perfectly horizontal. Determine the time the projectile is in motion, i.e. the time between its launching and its hitting the ground. Also determine its range, the distance from the bottom of the cliff to its point of impact with the ground. Finally, determine the displacement vector for the motion of the projectile

Explanation / Answer

Vx=30cos60=15 m/s Vy=30sin60=25.98 m/s Now time taken t1=2*vsin60/g t1=5.29 s also t2 is calculated by 100=25.98*t+1/2*9.81*t^2 t2=2.58 s total t=t1+t2=7.87 s ans b)horizontal distance S=15*7.87=118.05 m ans

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