Question

# A bullet is launched with an initial speed of 1245 m/s. The initial velocity in the...

A bullet is launched with an initial speed of 1245 m/s. The initial velocity in the y-direction is 515 m/s and the initial velocity in the x-direction is -1208 m/s. When the bullet reaches the peak of its trajectory, what is the velocity and the acceleration in the x-direction and what is the velocity and the acceleration in the y-direction? Consider up to be positive. Please include a diagram and an explanation of each step. Do you use the max height formula or some other kinematic formula?

Taking left side to be negative x dirn

The initial velocity U is 1245m/s

initial velocity in the x dirn is Ux -1208 m/s

The initial velocity in the y dirn is Uy 515 m/s

The angle of projectile is The time taken to reach max height is T . , g is the acceleration due to gravity

T = 52.6 s

So the final velocity in y comp is putting all the values we get

Vy = 0 m/s

The acceleration in the y comp is +g = 9.8 m/s2 downwards

The final velocity the x comp is

Vx =Ux = 1208 m/s as their is no acceleration in the x component

So the final acceleration in the x comp is 0m/s 2

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