Question

Cars Falls of a college and drops tour a ground in 0.5 second let's g equal to 10 metre per second -2 what is its speed on touching the ground what is average speed during 0.5 second how high is the lead from the ground

Answer 1

Explanation:

Given: time t=0.5s, acceleration a = 10m/s2

initial velocity u=0, final velocity v=?

$v = u + at$

$v = 0 + 10 \times 0.5 = 5m {s}^{ - }$

v=5m/s2

$s=ut+\frac{1}{2}×{at}^{2}$

$s=0×0.5+\frac{1}{2}×10×0.5×0.5 \\</p><p>=10×0.5×0.5×0.5=1.25m$

$average v=\frac{total\:displacement}{total\: time}\\=1.25/0.5=2.5m/{s}^{1}$