A particle is thrown vertically upward with initial velocity of 60 m/s. The distance covered by the particle in first 2 seconds of descent will be?

As v= u+at,
T=v-u/a=0-60/-10= 6sec
The distance covered
Upward speed,
S= u?t+1/2at^2
S=60?6-10?36
S=360-360
S=0
Time,t= 6-2=4sec
Down speed,
S=u?t+1/2at^2
After this add the values and then find total distance covered by adding upward speed and downward speed then subtarct upward speed and downward speed and you will get displacement.
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