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adelina 88 [10]

3 years ago

12

A Cannonball is shot at an angle of 35.0 degrees and is in flight for 11.0 seconds before hitting the ground at the same height

from which it was shot.
A. What is the magnitude of the inital velocity?B. What was the maximum height reached by the cannonball?C. How far, horizontally, did it travel?

1 answer:

nikdorinn [45]3 years ago

4 0

Answer:

Explanation:

According to Equations of Projectile motion :

$Time\ of\ Flight = \frac{2vsin(x)}{g}$

vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec

(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec

$Maximum Height = \frac{(vsinx)^{2} }{2g}$

(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m

$Horizontal Range = vcosx * t$

(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m

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