A projectile is launched from a platform 2 m above the ground at an angle of 38\

Question

A projectile is launched from a platform 2 m above the ground at an angle of 38' above the horizontal. The muzzle velocity of the projectile is 49 m/s. How far does the projectile go assuming it lands in a pit 4 m deep? How long is the flight time of the projectile? How fast is the projectile moving the moment before impact? What is the highest point the projectile reaches? How long does it take to get to the highest point? ACD has a diameter of 120 mm. Placed into a CD player it has a constant revolution speed of 1530 rpm. If the disc plays for 1.00 minute, how far has it rotated? When done playing the disc, the device is powered down. Due to friction, the disc decelerates at 0.777 m/sn2. How long does it take to come to a complete stop? How far does the disc rotate between the turning the device off and coming to a complete stop?

Explanation / Answer

(b)

Equation of motion in vertical direction:

-h = usinxt -0.5gt^2

2 = 49sin(38)xt -0.5x9.8xt^2

2 = 30.16t - 4.9t^2

4.9t^2-30.16t+2 =0

t = 6 seconds

C)

V = [ u^2 +2gh]^1/2

=[ 49^2 + 2x9.8x2]^1/2

= 49.39 m/s

d)

t' = 3.07

s = [(30.16+0)/2] x3.07

46.30 m

E)

time = 3.07 seconds[considering vertical motion]

2)

w = 160.22 rad/sec

a)

in one min disc makes 1530 revolution

=9613 radians

b)

a = -0.777

Apply

v = u+at

c)

using equation

s = ut +0.5at^2

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