a 50 g golf ball is driven A 50 g golfball is driven off a tee with a force of 4

Question

a 50 g golf ball is driven A 50 g golfball is driven off a tee with a force of 400N and contact time of 0.5 s. The golfball rises at an initial angle of 30 Degree from the horizontal. The ball lands straight down the Neglect air resistance and rotational motion of the ball (Equations of motion. Newton's Laws, energy and momentum) The initial speed of the golfball is m/s. The ball rises to height of m. The ball is in the air for s. The total machanical energy of the golfball in fight is j. The momentum change of the golfball at the instant it is driven kgm/s

Explanation / Answer

a) Impulse = F deltat = change in momentum

400 ( 0.5) = 0.050 ( v - 0)

v = 4000 m/s

b) initial vertical velocity = uy = vsin30 = 2000 m/s

it will rise until its velocity becomes zero.

using vf^2 - vi^2 = 2ad

0^2 - 2000^2 = 2(9.81)(H)

H = 203873.6 m

c) t = 2 v sin@ / g = 2 (4000)(sin30)/9.81 = 407.75 sec

d) total mechanical energy = (0.050)(4000^2)/2 = 400000 J

e) change in momentum = 0.050(4000) = 200 kg m/s

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