Question

The following two equations represent the x and y positions of a cannon ball after it is fired at an angle out of a cannon. Use these equations to determine the magnitude of the initial velocity of the cannon ball at t=0 s.  

x(t) = (1.2) t

y(t) = (2.7) t - (4.9) t²

Answer 1

Since there is no air resistance in horizontal direction, So x position of particle in projectile motion is given by:

x(t) = V0x*t

And since there is acceleration due to gravity in vertical direction, So y-position of particle in projectile motion is given by:

y(t) = V0y*t + (1/2)*a*t^2

g = -9.8

y(t) = V0y*t - 4.9*t^2

Now compare both equations:

x(t) = 1.2*t

y(t) = 2.7*t - (4.9)t^2

So,

V0x = 1.2 = Initial horizontal velocity

V0y = 2.7 = Initial vertical velocity

So Initial launching velocity will be:

V0 = sqrt (V0x^2 + V0y^2)

V0 = sqrt (1.2^2 + 2.7^2)

V0 = 2.95 m/s

Let me know if you've any query.