Question

a body of mass 5 kg is vertically thrown upwards with a speed of 10 M per second what is its kinetic energy find its potential energy at the highest point and also find the maximum height reached

Answer 1

m = 5 kg

u = 10m/s

K E = 1/2×m×u^2

= 1/2×5×100

= 250J

loss in KE = gain in PE

→ PE = 250

also PE = m×g×h

250=5×10×h

→ H = 5m

Answer 2

kinetic energy=0.5*5*10*10=250joulesl
we can find max. height by using third equation of motion

let max. height=s

${v}^{2} = {u}^{2} + 2as \\ {0}^{2} = {10}^{2} - 2(10)s \\ 0 = 100 - 20s \\ 20s = 100 \\ so \: s = 5m$
so max. height=5m
potential energy=mgh
so PE=5*10*5=250 joules

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