(a) With what velocity `(v_0)` should a ball be projected horizontally from the top of a tower so that the horizontal distance on the ground is `eta H`, where `H` is the height of the tower ?
(b) Also determine the speed of the ball when it reaches the ground.
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Options

A. `sqrt(((eta + 4)/(2)) g H)`
B. `sqrt(((eta^2 + 2)/(2)) g H)`
C. `sqrt(((eta^2 + 4)/(2)) g 2H)`
D. `sqrt(((eta^2 + 4)/(2)) g H)`

Answer

D

Explanation

(a) Given , horizontal distance = `eta H`, where `H` is the height of the tower.
To find out the sufficient velocity as asked, we have to use the formula : distance = velocity xx time. Here the time involved will be the time of flight as we are considering projectile motion,
which is given by `T = sqrt( 2 H) // g`
So, `v_0 T = eta H rArr v_0 sqrt((2 H))/(g)) = eta H rArr v_0 = eta sqrt((g H)/(2))`
(b) During the projectile motion, the horizontal component of velocity remains same and its vertical component keeps on changing under the effect of gravity. So horizontal speed, `v_x = v_0` and vertical speed, `v_y = sqrt(2 g H))`.
Total speed `= sqrt(v_x^2 + v_y^2))`
=`sqrt(v_0^2 + 2 g H))`
=`sqrt(eta^2 (g H)/(2) + 2 g H) = sqrt(((eta^2 + 4)/(2)) g H)`.