Question #342233

An object is being shot from a horizontal ground at an incline angle of 30 degree with respect to the ground at a speed of 49 m/s. Find the duration in seconds that the object is above the height of 13 m. Give your answer with one decimal place.


1
Expert's answer
2022-05-17T16:04:38-0400

The equation of motion of an object is given by

y=v0sinθtgt2/2y=49sin30t9.8t2/2y=24.5t4.9t2y=v_0\sin\theta t-gt^2/2\\ y=49\sin30^\circ t-9.8t^2/2\\ y=24.5t-4.9t^2

At the height 13 m we have

13=24.5t4.9t213=24.5t-4.9t^2

Roots of this equation

t1=0.60s,  t2=4.40st_1={0.60\:\rm s,}\;t_2=4.40\:\rm s

Hence, duration in seconds that the object is above the height of 13 m

Δt=t2t1=4.400.60=3.8s\Delta t=t_2-t_1\\ =4.40-0.60=3.8\:\rm s


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