QUESTION 1 A cannon fires a projectile with a speed $35 \mathrm{~m} / \mathrm{s}$ at an angle of $82^{0}$ up from the horizontal over level ground. How far (in meters) does the projectile fly?

Step 1 ：We are given a cannon fires a projectile with a speed of 35 m/s at an angle of 82 degrees up from the horizontal over level ground. We are asked to find how far (in meters) does the projectile fly.

Step 2 ：The range of a projectile is given by the formula: \[ R = \frac{v^2 \sin(2\theta)}{g} \] where: \(v\) is the initial velocity of the projectile, \(\theta\) is the angle at which the projectile is launched, and \(g\) is the acceleration due to gravity.

Step 3 ：In this case, \(v = 35 \, \text{m/s}\), \(\theta = 82^0\), and \(g = 9.8 \, \text{m/s}^2\).

Step 4 ：We can plug these values into the formula to find the range of the projectile.

Step 5 ：Doing the calculations, we find that the range of the projectile is approximately 34.45 meters.

Step 6 ：Final Answer: The projectile will fly approximately \(\boxed{34.45}\) meters.