A 20.0 kg mass moving at a velocity of + 3.0 m/s is stopped by a constant force of 15.0 n. how many seconds must the force act o
1 answer:
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Answer:
v = 46.99 m/s
Explanation:
The velocity of the ball just before it touches the ground, is given by the following formula:
(1)
vx: horizontal component of the velocity
vy: vertical component of the velocity
The vertical component vy is calculated by using the following formula:
(2)
vy: final velocity
voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)
g: gravitational acceleration = 9.8 m/s^2
h: height = 1.60m
You replace the values of the parameters in the equation (2):
vx is calculated by using the information about the horizontal range of the ball:
(3)
R: horizontal range of the ball = 20.0 m
You solve the previous equation for vo, the initial horizontal velocity:
The horizontal component of the velocity is constant in the complete trajectory, hence, you have that
vx = vo = 35 m/s
Finally, you replace the values of vx and vy in the equation (1):
The velocity of the ball just before it touches the ground is 46.99 m/s