A boy throws throws a ball vertically upward with speed of 25 m/s. on its way down it is caught is at a point 5 meters above the ground. How fast it was coming down at that point? How long did the trip take?
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No matter it is coming down or going up the magnitude of speed remains same at same height...
So apply 3rd equation of motion..
v^2 = U^ 2+2as
Where v is the velocity at height 5 m above ground..
U = 25 m/s
a = 10m/s^2
S = 5 m
V=?
So...
V^2 =( 25)^2 + 2*10*5
= 625 + 100
= 725
v = root(725)
calculate further approximate value and get ans...
aftet that....
Use
V = U +at
And find t..
So apply 3rd equation of motion..
v^2 = U^ 2+2as
Where v is the velocity at height 5 m above ground..
U = 25 m/s
a = 10m/s^2
S = 5 m
V=?
So...
V^2 =( 25)^2 + 2*10*5
= 625 + 100
= 725
v = root(725)
calculate further approximate value and get ans...
aftet that....
Use
V = U +at
And find t..
aisha200016:
what about the time?
After getting value of v... Use v =u +at and find time
ok thank you
U wlcm
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