A canon shell was projected on the valley, with a velocity of magnitude 1051.6 m

Question

A canon shell was projected on the valley, with a velocity of magnitude 1051.6 mi/h at an angle of 25.0 degrees above the horizontal. Ignore air resistance. The shell lands on the hill at an altitude of 1200 m. (a) Calculate the speed of the shell just before it strikes the ground. (b) Calculate the angle (in degrees) of the shell just before it strikes the ground (c) Calculate the horizontal distance from the launch point to the point where the shell strikes the ground (the range of the shell). (d) Calculate the maximum height of the shell over the valley. A canon shell was projected on the valley, with a velocity of magnitude 1051.6 mi/h at an angle of 25.0 degrees above the horizontal. Ignore air resistance. The shell lands on the hill at an altitude of 1200 m. (a) Calculate the speed of the shell just before it strikes the ground. (b) Calculate the angle (in degrees) of the shell just before it strikes the ground (c) Calculate the horizontal distance from the launch point to the point where the shell strikes the ground (the range of the shell). (d) Calculate the maximum height of the shell over the valley. (a) Calculate the speed of the shell just before it strikes the ground. (b) Calculate the angle (in degrees) of the shell just before it strikes the ground (c) Calculate the horizontal distance from the launch point to the point where the shell strikes the ground (the range of the shell). (d) Calculate the maximum height of the shell over the valley.

Explanation / Answer

Given that

Velocity=1051.6mi/h

Angle=25°

Altitude=1200m

Concept:

This belongs lo projectile motion

Formula:

Maximum height=u^2 sin(theta)/2g

Horizontal distance=u^2 sin 2(theta)/g

Vertical velocity=u sin(theta)

Calucalation:

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