A cannonball sits on top of a 10 meter cliff and is fired at a 10 degree angle w
ID: 2227897 • Letter: A
Question
A cannonball sits on top of a 10 meter cliff and is fired at a 10 degree angle with an initial velocity of 50 m/s. What is the maximum height of the cannonball relative to the ground? At maximum height what is the magnitude of the velocity? How long is the cannonball in the air?Explanation / Answer
upward component of velocity = 50 sin(10) = 8.68 m/s horizontal component of velocity = 50 cos(10) = 49.24 m/s At maximum height , upward velocity = 0. => 0^2 = (8.68)^2 - 2*(9.8)(h) => h = (8.68^2)/(19.6) => h = 3.844 meter SO, height wrt ground = (10+3.844) = 13.844 meter time taken to reach maximum height = (0-8.68)/(-9.8) = 0.8857 seconds => horizontal velocity = 49.24*0.8857 = 43.62 m/s since, at max height, the upward velocity is zero so, net magnitude of velocity at that point = 43.62 m/s net final displacement in upward direction = -10 meter ==> v^2 = 8.68^2 + 2*(-9.8)(-10) => v = -16.47 m/s i.e. 16.47 m/s downward => -16.47 = 8.68 - 9.8t => t = 2.566 seconds
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