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## Homework Statement

A cannon is inclined about the horizontal by an angle of 45° and it launches projectiles with initial velocity of ##v_p=300m/s##. A small airplane moves horizontally with a constant velocity of ##-100m/s##(so it is directed to the cannon) at high of ##500m##. If at ##t=0s##, the position of the airplane is ##7000m## away from the cannon, determine:

a) the time of which the projectile must be launch to hit the airplane.

b) what are the coordinates of the impact? at what time?

## Homework Equations

The result of point a) is wrong, I don't understand why?

## The Attempt at a Solution

starting with b)

From the system:

$$y=y_0+v_0sin(45°)-1/2gt^2$$

$$x=x_0+v_0cos(45°)$$

I found:

$$t=2.5s$$ and $$x=531m$$

so, the total time is $$t=\frac{x_(impact)-x_(airplane)}{v_(airplane)}64.7s$$

point a)

this should be ##64.7s-2.5s=62.2## because ##2.5s## is the time of the projectile to reach the coordinates ##(531m,500m)##.

Thanks for helping..

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