When a ball is thrown up vertically with velocity Vo it reaches a maximum height of h if one wishes to triple the maximum height then the ball should be thrown was the velocity
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case 1 :
Let the maximum height be H
Initial velocity: u
final velocity: v=0
using v^2= u^2+2gh
taking upward as positive and downward as negative. Since g acts downwards, it will be negative.
0= u^2-2gh
-u^2= -2gh
u^2/2g=h
u= V°
therefore V°^2/2g=h
Case 2
Height is tripled I.e 3h
using the same equation.
v^2= u^2 + 2gh
0= u^2-2g(3h)
substitute the value of h from case 1.
6gh= u^2
6g(V°^2/2g)= u^2
3V°^2= u^2
V°√3= u
Let the maximum height be H
Initial velocity: u
final velocity: v=0
using v^2= u^2+2gh
taking upward as positive and downward as negative. Since g acts downwards, it will be negative.
0= u^2-2gh
-u^2= -2gh
u^2/2g=h
u= V°
therefore V°^2/2g=h
Case 2
Height is tripled I.e 3h
using the same equation.
v^2= u^2 + 2gh
0= u^2-2g(3h)
substitute the value of h from case 1.
6gh= u^2
6g(V°^2/2g)= u^2
3V°^2= u^2
V°√3= u
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