projectile problem find distance, original velocity, how high did it go?

1)

Vm = muzzle velocity

Vh = Vm*cos(25)

Vvo = Vm*sin(25)

Tf = flight time = 48.4

t = time to climb to maximum altitude = flight time/2 = 48.4/2 = 24.2

Vv(t) = Vvo - gt = Vm*sin(25) - 9.8(24.2) => Vm = 561.2 m/s

Vh = Vm*cos(25) = 561.2*cos(25) = 508.6 m/s

Dh = VhTf = 508.6*48.4 = 24616 meters

2)

V = original velocity of ball

Vh = horizontal velocity of ball = Vcos43

Vvo = original vertical velocity of ball = Vsin43

Tf = flight time of ball = 2.3 sec

Df = flight distance of ball = 75 m

D = Vh*Tf = 75

Vh = Vcos43 = 75/2.3

a) V = 75/(2.3cos43) = 44.6 m/sec

b) V^2 - Vo^2 = 2ad

V = 0

Vo = Vvo = Vsin43

a = -g = -9.8

d = H = maximum height of the ball

0 - (V*sin43)^2 = 2*(-9.8)H

H = (44.6*sin43)^2/(2*9.8) = 47.2 m

enable the preliminary velocity be v0. The vertical area of which would be v0y = v0*cos? The projectile vertical velocity is vy(t) = v0y - g*t The projectile ascends till its vertical velocity is 0, so the ascent time is t = v0y/g the descent time is a similar, so complete flight time is T = 2*v0y/g you're given T (12.5 s), so clean up for v0y: v0y = T*g/2 you at present have the vertical velocity element, and you will discover the entire preliminary velocity from the eq v0y = v0*cos? (you're given ?) the gap it travels is the horizontal velocity element cases the flight time. The horizontal velocity element is v0x = v0*sin? D = v0x*T = v0*sin?*T (you have already calculated v0. The vertical distance traveled is ymax = v0y*ta - 0.5*g*ta² t is the ascent time calculated till now, ta = v0y/g ymax =0.5* v0y²/g