a laboratory projectile launcher is set at zero degrees elevation and secures at

Question

a laboratory projectile launcher is set at zero degrees elevation and secures at a high of 1.25 meters above the floor. the projectile is launched and lands on the floor 3.15 meters away. determine where the projectile will land if the launcher is angled at thirty seven degrees. (assume the Sy of the launcher muzzle is also raised by 4 cm) a laboratory projectile launcher is set at zero degrees elevation and secures at a high of 1.25 meters above the floor. the projectile is launched and lands on the floor 3.15 meters away. determine where the projectile will land if the launcher is angled at thirty seven degrees. (assume the Sy of the launcher muzzle is also raised by 4 cm) a laboratory projectile launcher is set at zero degrees elevation and secures at a high of 1.25 meters above the floor. the projectile is launched and lands on the floor 3.15 meters away. determine where the projectile will land if the launcher is angled at thirty seven degrees. (assume the Sy of the launcher muzzle is also raised by 4 cm)

Explanation / Answer

Here ,

let the initial speed of projectile is u m/s

time of flight , tf = sqrt(2*h/g)

tf = sqrt(2 * 1.25/9.8)

tf = 0.5051 s

Now, speed of launch , u = 3.15/0.5051

u = 6.24 m/s

>>for theta = 37 degree

Height , h = 1.25 + 0.04 = 1.29 m

using second equation of motion

-1.29 = 6.24 * sin(37) * t - 0.5 * 9.8 * t^2

solving for t

t = 1.02 s

Now , distance from gun = 6.24 * cos(37) * 1.02

distance from gun = 5.08 m

the projectile will land 5.08 m from the gun

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