Check of my answers for projectile motion please

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SUMMARY

The discussion revolves around calculating projectile motion for a pilot dropping a bomb from a height of 98 meters while flying at a speed of 45 m/s. The time taken for the bomb to hit the ground is accurately calculated as 4.47 seconds using the formula t = √(2h/g). The horizontal distance for the bomb release is determined to be 201 meters. However, the initial margin of error calculation of 3 seconds is incorrect; the correct margin of error is approximately 0.33 seconds, which corresponds to a horizontal distance of 15 meters, the diameter of the target.

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ghostbuster25
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Homework Statement


ok the question is as follows. A pilot flying at 45m/s is trying to bomb a target on the ground 15m in diameter. he is at a height of 98m
(a)how long does it tke for the bomb to hit the ground?
(b)at what horizontal distance does the pilot have to be to release the bomb?
(c)what margin of error does he have in hitting the target?

Just want a check on my solutions
(a)
t=[tex]\sqrt{}2h/g[/tex] will give me [tex]\sqrt{}2*98/9.8[/tex] = 4.47secs

(b)
R=u[tex]\sqrt{}2h/g[/tex] will give me 45m/s[tex]\sqrt{}2*98/9.8[/tex] = 201m

(c) 45m/s / 15 = 3 seconds of error

seems resonable to me!?

If I am wrong please show me where

Thanks
 
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Parts a and b are OK. Your answer to part c does not look right. Are you saying that he can release the bomb 3 seconds late and still hit the target? In three seconds he travels 135 m and that's way too much. What you need is the time required to travel the length of the target, i.e. 15 m. If he travels farther than that, he misses.
 
ah yer that makes more sense, 15m/45m/s = 0.3333333 s
 

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