Respuesta :
Answer: 9.3m/s
Explanation:
Your question isn't complete but let me help out:
A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball. What is the magnitude of the velocity after it hits the ground?
We would use Newton's law of motion to solve this which goes thus:
F = ma
f = m(v-u)/t
Cross multiply
ft = m(v-u)
where,
f = 55
t = 45/1000 = 0.045
m = 0.0060
u = -32
v = Unknown
Therefore,
55 × 45/1000 = 0.060(v - -32)
55 × 0.045 = 0.060(v + 32)
2.475 = 0.06(v + 32)
2.475 = 0.06v + 1.92
0.06v = 2.475 - 1.92
0.06v = 0.555
v = 0.555/0.06
v = 9.25m/s
v = 9.3m/s Approximately
