Suppose that you use a launcher with a muzzle velocity of 150 m/s. The launcher

Question

Suppose that you use a launcher with a muzzle velocity of 150 m/s. The launcher is tilted at an angle of 38 degrees relative to the ground, and a target is located some distance away, at an elevation 57 m BELOW the launcher. Calculate the following quantities: A) the initial horizontal and vertical velocities of the projectile B) the horizontal distance Xt that the launcher should be positioned relative to the target in order for the projectile to hit the target. Use equation: 1+ sqrt1-2BetaYt/ Beta(tan(angle)) C) The "hang time" of the projectile before it hits the target. Use equation: 1+ sqrt1-2BetaYt/BetaVy0 D) The projectile's horizontal, vertical, and total velocity upon hitting the target.

Explanation / Answer

A) Initial horizontal velocities of the projectile

V0x = u * cos theta = 150 * cos 38

= 118.20 m/s

Initial verticle velocities of the projectile.

V0y = u * sin theta = 150 * sin 38

= 92.35 m/s

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B) yt = 57 m

V0y = Initial vertical velocity = 92.35 m/s

xt = [1+ sqrt( 1-(2 * 9.8 * 57))] / (9.8 * Tan (38))

= 0.1306 m

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C) t = [1+ sqrt( 1-(2*9.8*57)) ] / (9.8* 92.35)

= 0.0011 s

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D) Projectile horizontal velocity = V0x = 118.2 m/s

Projectile vertical velocity = 0 m/s

Projectile total velocity = V0x i + Vy j

= 118.2 m/s

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