Answer:
26.2 m/s
Explanation:
We can find the speed of the cannonball just by analyzing its vertical motion. In fact, the initial vertical velocity is
[tex]u_y = u sin \theta[/tex] (1)
where u is the initial speed and [tex]\theta = 45.0^{\circ}[/tex] is the angle of projection.
We can therefore use the following suvat equation for the vertical motion of the ball:
[tex]v_y = u_y + at[/tex]
where [tex]v_y[/tex] is the vertical velocity at time t, and [tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity. The time of flight is 3.78 s, so we know that the ball reaches its maximum height at half this time:
[tex]t=\frac{3.78}{2}=1.89 s[/tex]
And at the maximum height, the vertical velocity is zero:
[tex]v_y=0[/tex]
Substituting these values, we find the initial vertical velocity:
[tex]u_y = v_y - at = 0-(-9.8)(1.89)=18.5 m/s[/tex]
And using eq.(1) we now find the initial speed:
[tex]u=\frac{u_y}{sin \theta}=\frac{18.5}{sin 45.0^{\circ}}=26.2 m/s[/tex]
