Problem 79.

The highest barrier that a projectile can clear is 20 m, when the projectile is launched at an angle of 40.0 degrees above the horizontal. What is the projectile's launch speed?

Solution:

The meaning that the highest barrier the projectile can clear is 20 m is that the maximum height of the projectile is 20 m.

Since we are interested only in the height of the projectile then we can consider only the motion of the projectile along the vertical direction (axis y). Let us introduce the launch speed of the projectile as . Then the y component of the initial velocity of the projectile is

................................................(1)

The motion along axis y is a free fall motion and it is described by the following equations (only two equations are independent):

We know the initial height (y-coordinate) of the projectile: .

Then we introduce the final point – the point at which the projectile has the maximum height. We know this height: .

We also know that the y-component of final velocity (at the maximum height) is zero. We substitute these values in the third equation and obtain

From this equation we can find :

Then from equation (1) we can find the launch speed of the projectile: