Calculate (a) the max. Height reached by bullet above the ground. (b) Total time of flight (c) Velocity with which the bullet hits the ground.

Let A be the intial & B be the highest point reached by bullet

x be the height of B from A t be total time of flight

Now, considering vertical motion of bullet From A to B,

$S=x $

V = 0 (at highest point vertical velocity is 0)

$U= 100 \sin50 \\ A = -9.81 $

Applying, $V2 = U2 + 2AS \\ 0 = (100 \sin50)2 - 2(-9.81)x \\ x = 299.095 m $

Hence, maximum height reached by bullet is $299.095 + 300 = 99.095m $

Now, considering vertical motion from A to C

$S = -300 \\ U = 100 \sin50 $

V = vY (Y component of velocity at C)

T = t (total time of flight)

$A = -9.81 $

Now, applying $S = UT + (1/2)AT2 \\ -300 = (100 \sin50)t - (1/2)(-9.81)t2 $

Solving quadratic in C, we get

$t = 18.86 m/s $

Applying $V = U + AT \\ vY = 100 \sin50 - 9.81(t) \\ vY = - 108.412 m/s = 108.412 m/s $

Also, as there is no accleration acting in x direction VX remains same

Hence at $C, vX = 100 \cos50 = 64.28 m/s$ & $vY = 108.412 m/s $

$$ V_C = \sqrt{( V_x^2 + V_y^2) }= 126.035 m/s \searrow$$

Angle made by velocity with horizontal $= \tan^{-1}(vY/ vX) = 59.33^\circ $

The max Height reached by bullet above the ground is $599.095 m$. Total time of flight is $18.86 sec.$ and Velocity with which the bullet hits the ground is $126.035 m/s \searrow @ \theta = 59.33^\circ$