From the top of the building, Sam threw a 0.06 kg ball at an angle such that the

Question

From the top of the building, Sam threw a 0.06 kg ball at an angle such that the upward initial velocity is 4.0 m/s and the horizontal initial velocity is 8.0 m/s. The height of the building is 99 m. Using 10 N/kg for the magnitude of g, the upward direction as positive, and assuming that the ball does not reach ground,

a) what is in m/s the horizontal component of the velocity of the ball 0.50 seconds after it was thrown?

b) what is in m/s the vertical component of the velocity of the ball 0.50 seconds after it was thrown?

c) what is in m/s2 the vertical component of the acceleration of the ball 0.50 seconds after it was thrown?

d) how many forces are acting on the ball 0.50 seconds after it was thrown?

e) what is in Newtons the net force acting on the ball 0.50 seconds after it was thrown?

Explanation / Answer

part a)

since there is no acceleration in x direction

so horizontal componenet of velocityremain the same after 0.5 sec that is 8 m/s

part b)

vertical componenet of velocity

v=u+at=4-10(0.5)=4-5=-1 m/s, that is the ball is coming downwards

part c)

the verticle componenet of acceleration will remain constant

a=10 m/s2

because a is constant

part d)

accleration a is only in one direction that is downwards,

so force is acting only in one direction, that is downwards

part e)

F=ma

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