Question

In: Physics

During a fireworks display, a shell is shot into the air with an initial speed of...

During a fireworks display, a shell is shot into the air with an initial speed of 91.55 m/s at an angle of theta degrees above the horizontal. The fuse is timed to 8.21 seconds for the shell just as it reaches its highest point above the ground. If the horizontal displacement of the shell is 149.68 meters when it explodes, what is the height (in unit of meters) when the shell explodes?

You must use g = 9.8 m/s².

Expert Solution

Time taken by the shell to explode,

Initial launch velocity,

Horizontal displacement of the shell when it explodes,

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Consider the horizontal motion of the shell

There is no acceleration in the horizontal direction, so horizontal velocity remains constant.

Horizontal velocity = Horizontal distance/time

ucos =x/t

x/(u*t)

149.68m/(91.55m/s*8.21s)

0.1991

78.513 deg

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Consider the vertical motion of the shell

Use formula

y = usin *t - 0.5*gt²

y = 91.55m/s*sin78.513 * 8.21s - 0.5*9.8m/s²*(8.21s)²

y = 91.55*sin78.513 * 8.21 - 0.5*9.8*8.21²

ANSWER: y=406.29m

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genius_generous answered 6 months ago

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