a ball thrown up is caught by the thrower after 4s. how high did it go and with what velocity was it thrown ? gow far the highest point 3s after it was thrown?
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For half journey..[throwing up]
t =2 sec ( time of throwing up= time of comming down)
g= -10m/s^2
v=O
S= ?
a=v-u/t
-10=0-u/4
-40=-u
so, u = 40m/s
NOW finding s by
2as = v^2 - u^2
-20s = 0-1600
S = -1600/-20
S = 80m
Total dis. travelled : 160m
Distance travelled in 3sec
t=3sec
g=-10
v=0
S=?
a = v-u/t
-10=0-u/3
-30=-u
u = 30m/s
Now finding s by 2as= v^2-u^2
2*-10*s = 0-900
-20s=-900
S = 900/20
S=45m
hope it helps
t =2 sec ( time of throwing up= time of comming down)
g= -10m/s^2
v=O
S= ?
a=v-u/t
-10=0-u/4
-40=-u
so, u = 40m/s
NOW finding s by
2as = v^2 - u^2
-20s = 0-1600
S = -1600/-20
S = 80m
Total dis. travelled : 160m
Distance travelled in 3sec
t=3sec
g=-10
v=0
S=?
a = v-u/t
-10=0-u/3
-30=-u
u = 30m/s
Now finding s by 2as= v^2-u^2
2*-10*s = 0-900
-20s=-900
S = 900/20
S=45m
hope it helps
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