Asked by Sharma
a body of mass 5 kg is thrown vertically upwards with a speed of 10m/s what is its k.e. when it is thrown find its potential energy when it reaches at highest poiint also find maximum height attained by the body?
Answers
Answered by
Henry
a. KE = 0.5m*Vo = 2.5*(10)^2 = 250 Joules
b. hmax = (Vf^2-Vo^2)/2g.
hmax = (0-(10)^2 / -19.6 = 5.10 m.
c. PE = mg*h = 5 * 9.8 * 5.1=250 Joules
b. hmax = (Vf^2-Vo^2)/2g.
hmax = (0-(10)^2 / -19.6 = 5.10 m.
c. PE = mg*h = 5 * 9.8 * 5.1=250 Joules
Answered by
Adiba
m=5kg
V=10m/s
K.E=1/2mv^2
P.E=mgh
K.E=P.E
K.E=1/2mv^2
=1/2x10x5x5
=125
P.E=125
Mgh =P.E
5x10xh=125
10h=125/5=25
h=25/10=2.5 answer
V=10m/s
K.E=1/2mv^2
P.E=mgh
K.E=P.E
K.E=1/2mv^2
=1/2x10x5x5
=125
P.E=125
Mgh =P.E
5x10xh=125
10h=125/5=25
h=25/10=2.5 answer
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