A 3.00 kg projectile is launched from ground level traveling at 72.6 m/s and dir

Question

A 3.00 kg projectile is launched from ground level traveling at 72.6 m/s and directed at an angle of 67.3° above horizontal. The ground is perfectly horizontal and air- resistance is negligible.

1. a) What is the initial velocity of the projectile in component form?

2. b) What is the magnitude of the projectile’s velocity when it is at its highest point?

3. c) At what TWO values of time is the projectile located at a height of 44.1 m above the ground?

4. d) Relative to where it was launched from, what is the position of the projectile 7.80 seconds after being launched?

Explanation / Answer

a) vi = vix + viy = (72.6cos67.3) j + (72.6sin67.3)j = (28.02)i+(67.0)j m/s

b) At highest point viy=0m/s

and vix = vfx = (28.02)i m/s

Thus, vf = vfx = (28.02)i m/s

|vf| = 28.02 m/s

c) y = viy*t -1/2gt^2

44.1 = 67.0*t -1/2*9.8*t^2    => t= 0.7s or t= 13.0s

d) y= viy*t -1/2gt^2 = 67.0*7.08 -1/2*9.8*7.08^2 = 228.74 m

x = vix*t = 28.02*7.08 = 198.38 m

d = 198.38 i + 228.74 j

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