A bullet is fired at an angle of 60 degrees with respect to the horizontal with

Question

A bullet is fired at an angle of 60 degrees with respect to the horizontal with an initial speed of 120m/s. How high up does it go and how long does it take to reach the maximum height? How long does it take to reach the ground and what horizontal distance does it travel when it reaches the ground? What is its velocity when it reaches the ground? Draw a picture of the velocity when it reaches the ground. What is its velocity and position 15seconds after it is launched? A block is of mass 4kg is moving at 200m/s and is projected up an inline plane

Explanation / Answer

Here,

theta = 60 degree

initial speed , u = 120 m/s

let the maximum height is h

h = (v * sin(theta))^2/(2 * g)

h = (120 * sin(60))^2/(2 * 9.8)

h = 551 m

the maximum height reached by bullet is 551 m

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time taken to reach ground = 2 * v * sin(theta)/g

time taken to reach ground = 2 * 120 * sin(60)/9.8

time taken to reach ground = 21.21 s

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horizontal distance travelled by bullet = v^2 * sin(2 * theta)/g

horizontal distance travelled by bullet = 120^2 * sin(2 * 60)/9.8

horizontal distance travelled by bullet = 1272 m

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let the final speed of bullet is same as the initial speed of bullet

final speed of bullet = 120 m/s

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