A particle is thrown upward with a velocity of 50m/sec .what will be the velocity at its highest point of the journey. How high the particle rise. What time would it taken to reach the highest point (g=10m/sec)

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Solution

U= 50m/s
$g = 10 \frac{m}{s^{2}}$

The velocity at the highest point will be 0 m/s
Thus v=0

Height can be calculated by
$v^{2} - u^{2} = 2 g s$
-20s= -2500
s= 125m

Time :
v= u+gt
0=50-10t
-50=-10t
t= 5sec

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A particle is thrown upward with a velocity of 50m/sec .what will be the velocity at its highest point of the journey. How high the particle rise. What time would it taken to reach the highest point (g=10m/sec)

U= 50m/sg=10ms2 The velocity at the highest point will be 0 m/s Thus v=0 Height can be calculated byv2−u2=2gs -20s= -2500 s= 125m Time : v= u+gt 0=50-10t -50=-10t t= 5sec

U= 50m/s
$g = 10 \frac{m}{s^{2}}$

The velocity at the highest point will be 0 m/s
Thus v=0

Height can be calculated by
$v^{2} - u^{2} = 2 g s$
-20s= -2500
s= 125m

Time :
v= u+gt
0=50-10t
-50=-10t
t= 5sec