A ball is thrown vertically upwards with a velocity of 49 m/s. calculate: i. the max height to which it rises. (ii) total time it takes to return to surface of the earth.
Answers
Answered by
2
u = 49 m/s
g=10m/s^2
v=0
v=u +at
= 49 +10t =0
10t = 49
t =4.9 sec
2as =v^_u^
2×-10×s =-2401
-20s = -2401
s =2401/20
s =120.05
g=10m/s^2
v=0
v=u +at
= 49 +10t =0
10t = 49
t =4.9 sec
2as =v^_u^
2×-10×s =-2401
-20s = -2401
s =2401/20
s =120.05
Answered by
1
hi.
it's_ur_ansr...
u=49m/s
g=10 m/sec^2
v=0
1:-) v=u+at
10t=49
t=4.9 sec
2:-) 2as=v^2-u^2
-20s = -2401
s=120.05.
hope helps uh @Princə
it's_ur_ansr...
u=49m/s
g=10 m/sec^2
v=0
1:-) v=u+at
10t=49
t=4.9 sec
2:-) 2as=v^2-u^2
-20s = -2401
s=120.05.
hope helps uh @Princə
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