Question
If the initial velocity is Vo, the launch angle is , and the max height is H, show that the maximum heightwhich the ball rises is H=1/2(Vo^2sin^2 /g) And If a projectil is launched horizontally from height y and its triangle x as shown in the figure below, show thatthe initial velocity is given by Vo= ( x)^2g/2( y) If the initial velocity is Vo, the launch angle is , and the max height is H, show that the maximum heightwhich the ball rises is H=1/2(Vo^2sin^2 /g) And If a projectil is launched horizontally from height y and its triangle x as shown in the figure below, show thatthe initial velocity is given by Vo= ( x)^2g/2( y)
Explanation / Answer
In vertical direction : -------------------- Initial velocity u = V o sin accleration a = - g distance travelled S = H final velocity V = 0 from the relation V ^2 - u ^ 2 = 2aS - ( V o sin ) ^ 2 = -2gH from this maximum height H = ( Vo sin ) 2 / 2g = ( 1/ 2) V o2 sin 2 /g (b).Time taken to reach the ground t = [ 2y / g ] we know range x = initial velocotiy * time from this initial velocity v o = x /time = x [ g / 2 y ] = (x)^2g/2(y) (b).Time taken to reach the ground t = [ 2y / g ] we know range x = initial velocotiy * time from this initial velocity v o = x /time = x [ g / 2 y ] = (x)^2g/2(y)