Projectile Motion A person shoots a gun toward a wall which is 45m away in horiz

Question

Projectile Motion A person shoots a gun toward a wall which is 45m away in horizontal direction. The release point of the bullet is 1.55m above the ground, and the initial velocity of the bullet encloses an angle of 4 degree with the horizontal. The bullet hits the wall after 0.4s. What are x and y component of the initial velocity of the bullet? How high above the ground does the bullet hit the wall? What is the angle of the velocity vector relative to the horizontal just before hitting the wall?

Explanation / Answer

ucos(theta) is the horizontal velocity of the bullet.

So u cos(theta) * (0.4) = 45m

So u = 45/0.4*(cos( 4 degrees)) = 112.8m/s m/s

Soy component will be 112.8 * sin(4degrees) = 7.869 m/s

S = ut - 1/2 gt^2

So 7.869 * 0.4 - 4.9 * (0.4)^2 = 2.636 m

So Total height = 1.55 + 2.636 = 4.196m

At time of hit vertical velocity will be u-gt = 7.869 - 9.8 *0.4 = 3.949.

So Tan (alpha) = 3.949/112.5

alpha = 2.01 degrees

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