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.


Expert's answer

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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

LATEST TUTORIALS
APPROVED BY CLIENTS