Question

a projectile is fired horizontally from the top of a tower find the expression for its time of descent and horizontal range

Answer 1

it is designed only in the degree of 45 degree

Answer 2

Answer:

Explanation:

Let height of tower is h.

Along vertically downward , we have:-

u (along downward) = 0

a = g and s = h

So,

 s = 1/2 at square (∵ u = 0)

$t^{2}$ = 2s/a = 2h/g  (since s =h and a = g)

so, t = $\sqrt{2h/g}$  (Ans.)

Now, along horizontal,

 initial velocity = u

a = 0

so, s = ut (since a =0  so, 1/2 a$t^{2}$ = 0)

So, horizontal range,R = u$\sqrt{2h/g}$       (ans.)