<span>braking and returning suddenly to the roadway</span>
Let’s do this together!
Okay so the acceleration formula is vf-vi over time .
So the initial velocity (vi) 7m/s final velocity (vf) is 16m/s so we’re going to subtract 16-7 which is 9
M/s
So the time is 5s so 9m/s divided into 5s is 1.8m/s/2
So the answer is 1.8m\s2
Answer:
-39.2m/s
Explanation:
Using the equation of motion;
v = u + at
Since the ball is thrown upward, the acceleration due to gravity acting on it will be negative, hence a = -g
v = u - gt
Since g = 9.8m/s²
t = 4.0s
u = 0m/s
v = 0 + (-9.8)(4)
v = 0 + (-9.8)(4)
v = -39.2m/s
Hence the speed of the ball before release is -39.2m/s
W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s
