(a) A projectile with mass 5 kg is launched horizontally from a height of 10 m a
ID: 1461568 • Letter: #
Question
(a) A projectile with mass 5 kg is launched horizontally from a height of 10 m and the range is measured 40 m. (a) What is the launch speed? (b) What is the angle the projectile hits the ground at? (c) What is the KE when it hits the ground? (b) A rifle has mass of 4.5 kg and it fires a bullet of mass 10 g at a speed of 820m/s. What is the recoil speed of the rifle as the bullet leaves the gun barrel? (c) A ball of mass 2.0 kg moving with a weed of 3m/s hits a wall and bounces bock wits the 80% of the initial speed. What is the change in KE? What type of collision is this?Explanation / Answer
A]
a] Range, R = u*sqrt[ 2h / g ]
40m = u*sqrt [ 2*10m / 9.8m/s2]
u = 40/1.4286
u = 28m/s
b] angle = tan-1[g*t / u]
't' is the time of descent given by
t = sqrt [ 2h / g ] = sqrt [ 2*10 / 9.8 ]
t = 1.4286 s
therefore,
angle = tan-1[9.8*1.4286 / 28]
angle = tan-1[0.5]
angle = 26o33'54''
c] KE = 1/2*m*v2
where, 'v' is the final velocity when it reaches the ground; given by
v = sqrt [ u2 + g2t2] = sqrt [ 282 + (9.82*1.432) ] = sqrt [ 784+(96.04*2.05)] = sqrt [980.4]
v = 31.31 m/s
KE = 0.5*5*31.312
KE = 2450.98 J
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