To find the maximum and minimum velocities that the ball can have and still land in the trunk of the truck, we need to use the equations of motion for the horizontal and vertical positions of the ball as a function of time. Since the truck is moving away from the ball, the horizontal position of the ball will be given by:
x(t) = x0 + Vt - v0cos(θ)*t
Where x0 is the initial horizontal distance from the back of the truck to the ball, V is the velocity of the truck, t is the time, and v0 is the initial velocity of the ball.
Similarly, the vertical position of the ball will be given by:
y(t) = y0 + v0*sin(θ)*t - (1/2)gt^2
Where y0 is the initial height of the ball, v0 is the initial velocity of the ball, g is the acceleration due to gravity and t is the time.
We know that the ball will land in the trunk of the truck if the horizontal position of the ball is equal to the length of the trunk (L) at the same time that the vertical position of the ball is equal to the initial height of the ball (y0).
We can find the time of flight by setting y = 0,
t = (2v0sin(θ))/g
Now, we substitute this value of t and y0 = 0 in the equation of x(t) and equate it with L.
x0 + Vt - v0cos(θ)*t = L
We can solve this equation for v0 to get the initial velocity of the ball.
v0 = (L + x0cos(θ) - Vt)/(cos(θ) - t*sin(θ))
Finally, we know that the minimum and maximum initial velocities will correspond to the maximum and minimum values of sin(θ) and cos(θ). Since θ = 45°, we can find the maximum and minimum velocities as follows:
v0_max = (L + x0cos(45) - Vt)/(cos(45) - tsin(45))
v0_min = (L + x0cos(45) - Vt)/(cos(45) + tsin(45))
Note that in this case x0=5m, L=2.5m, θ=45°, V=9m/s, t = (2v0sin(45))/g and g is 9.8m/s^2
you can use these values to find out v0_max and v0_min.