How long will it take a shell fired from a cliff at an initial velocity of 800 m/s at an angle 30 degrees below the horizontal to reach the ground 150m below?

The shell will take about 0.34 seconds to reach the ground.

T is given as 800 m/s. The factors that would impact time are gravity and angle.

The factor of gravity (g) is a constant given as = 9.81. Any object on earth will accelerate at the speed of 9.81 meters per second if they are in a free fall.

Hence, T = 800Sin(30°) / g = 800 * 0.5

This is equals   400/9.81 = 40.773 or 40.78 seconds.

Given that the vertical velocity of the shell is zero because it is accelerating at an angle towards the earth, it will rise at an average speed that can be computed as:

1/2 (800sin30°) = 1/2 * 400 = 200m/s for Tsec and arise of H which is 8,156m.

The shell having risen falls from rest at the height given above.

S therefore,  = 150 +H, that is 150 + 8,156m. This is equals to 8,306m.

Next, we compute for the Kinematics equation of distance in relation to time (t) where the initial speed is 0.

S = 1/2(gt)²

thus 8,306 = 0.5 * 9.81*t² with elementary cross multiplication and the  and the squaring of both sides, we have

t =

t= 41.14608 which is approximately 41.12 seconds.

Because the time taken for the shell to rise and then to fall to the ground is the sum of T and t. the time taken to fall or the time taken for the shell to travel towards to the ground is:

t - T = 41.12 - 40.78

= 0.34 seconds

See the link below for more about Velocity:

link

answer; how far out of place which the object finished relative to the starting location;