On level ground a shell is fired with an initial velocity of 50.0 m/s at 64.0 ab

Question

On level ground a shell is fired with an initial velocity of 50.0 m/s at 64.0 above the horizontal and feels no appreciable air resistance.

Find the horizontal and vertical components of the shell's initial velocity.

How long does it take the shell to reach its highest point?

Find its maximum height above the ground.

How far from its firing point does the shell land?

At its highest point, find the horizontal and vertical components of its acceleration.

At its highest point, find the horizontal and vertical components of its velocity.

Explanation / Answer

Here ,

initial velocity , u = 50 m/s

theta = 64 degree

For the horizontal and vertical components of initial velocity

horizontal initial velocity = u * cos(theta)

horizontal initial velocity = 50 * cos(64) m/s

horizontal initial velocity = 21.9 m/s

vertical initial velocity = u * sin(theta)

vertical initial velocity = 50 * sin(64) m/s

vertical initial velocity = 44.94 m/s

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let the time taken is t

time taken = vy/g

time taken = 44.94/9.8

time taken = 4.59 s

the time taken to reach the highest point is 4.59 s

-----------------------------------------

maximum height = vy^2/(2 * g)

maximum height = 44.94^2/(2 * 9.8)

maximum height = 103.04 m

the maximum height above ground is 103.04 m

--------------------------------------

distance from launch point = vx * 2 * t

distance from launch point = 2 * 4.59 * 21.9 m

distance from launch point = 201.04 m

-----------------------------------------

at the highest point

horizontal acceleration = 0 m/s^2

vertical acceleration = 9.8 m/s^2 downwards

----------------------------

at the highest point

hoziontal velocity = 21.9 m/s

vertical velocity = 0 m/s

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