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a particle is projected vertically up with a speed of 10 m/s . find the times after which it will pass through a point 3.2m above the point of projection . (g=10 m/s)

keerthi , 8 Years ago
Grade 11
anser 3 Answers
Mohd Mujtaba
Here u=10;h=3.2;g=10.. so using 2 law of motion s=ut-1/2gt^2 as a=-gPut value you get t=0.4,1.6two answer is because if you throw upward ball it passes through two times at a given point .Thank
Last Activity: 8 Years ago
shreya shukla
u=10m/s
s=3.2m
g= -10m/s^2
now according to the question we first have to find final velocity at point 3.2m above the ground
v^2-u^2= -2gs
v^2-100=-64
v=+6,-6m/s we choose +ve sign as velocity is in above direction
now apply eq. v=u-gt
6=10-10t
t=0.4s
Last Activity: 8 Years ago
krishna bhagat
Given we have
u=10m/s
g= -10m/s^2
now ,
v^2-u^2= -2gs
v^2-100=-64 (s=3.2)
V=6m per s.
v=u-gt
6=10-10t
t=0.4 second
t=0.4s
Last Activity: 7 Years ago
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