What is the final speed right before it strikes the floor?

Question

A projectile launched at 140.0 m/s at 40 degree angle. What is the final speed right before it strikes the floor.

Answer 1

initial velocity = 140 m/s

Angle = 40 deg

g = 9.8 m/s²

t = (2x140xsin40)/9.8 = 18.365 secs

only Y- component of velocity will change and x-component will be constant, therefore;

Vy = Vsin(theta) - gt

V_y = [140xsin(40)] -(9.8*18.365) = - 89.99 m/s

negative sign shows that it is downward.

V = [(V_x²+V_y²)]^{^1/2} = [(140xCos40)²+(89.99)²]^{^1/2}

V = 140 m/s (this is the final speed before it hits the ground)

Angle = tan^-1(V_y/V_x)

Angle = tan^-1[89.99/(140xCos40)] = 40 deg