A ball is rolled up a constant slope. After 3.6s it reaches its maximum displacement of 2.6m and then begins to roll back down. What was the initial velocity of the ball when it started up the slope
4 years ago
Answered By Majid B
t=3.6 s
d=2.6 m
$v_f$vƒ =0
$v_i$vi =?
a = constant => $d=\frac{\left(v_i+v_f\right)}{2}\times t$d=(vi+vƒ)2×t => $v_i=\frac{2d}{t}-v_f=\frac{2\times2.6}{3.6}-0=1.44$vi=2dt−vƒ=2×2.63.6−0=1.44 m/s
4 years ago
Answered By Sosimo H
S=1/2(V0 + vf ) t
2S = V0 t + Vf t ; V0 t= 2S - Vfr t
Vf = 0 , Therefore Vo = 2S/ t ; V0= 2(2.6)m/ 3.6 s= 1.44 m/s
4 years ago
Answered By Majid B
t=3.6 s
d=2.6 m
$v_f$vƒ =0
$v_i$vi =?
a = constant => $d=\frac{\left(v_i+v_f\right)}{2}\times t$d=(vi+vƒ )2 ×t => $v_i=\frac{2d}{t}-v_f=\frac{2\times2.6}{3.6}-0=1.44$vi=2dt −vƒ =2×2.63.6 −0=1.44 m/s
4 years ago
Answered By Sosimo H
S=1/2(V0 + vf ) t
2S = V0 t + Vf t ; V0 t= 2S - Vfr t
Vf = 0 , Therefore Vo = 2S/ t ; V0= 2(2.6)m/ 3.6 s= 1.44 m/s
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