College physics?

Question

The invention of the cannon in the fourteenth century made the catapult unnecessary and ended the safety of castle walls. Stone walls were no match for balls shot from cannons. Suppose a cannonball is launched from a height of 1.10 m, at an angle of elevation of 30.0° with an initial velocity of 50.0 m/s, toward a castle wall. (a) The range of a projectile is defined as the horizontal distance traveled when the projectile returns to its original height. Derive an equation for the range in terms of v0, g, and angle of elevation q. (b) What will be the range reached by the projectile, if it is not intercepted by the wall?

Answer 1

The Vector Equation for Ideal Motion is r = [(v0cosa)t + rocosa]i + [(v0sina)t - (1/2)gt2 + rocosa]j

Answer 2

The two formulas for motion under constant acceelration are:

Vf = Vi + A T

D = Vi T + (1/2) A T^2

where:

T is the time interval,

A is the acceleration (= g)

Vi is the initial velocity

Vf is the final velocity

D is the distance travelled in the interval.

In this case, these equations only apply to the vertical; the horizontal velocity is constant.

So, using the initial velocity v0 and the initial angle, you can compute Vh and Vi.

The cannonball keeps rising until its vertical velocity = 0. So you can use the velocity equation to determine how long this takes. It takes as long to fall the same distance as to took to rise, so this gives you the time till the projectile reaches its original height.

The horizontal velocity is constant so velocity x times = distance.

I'm not sure what part b is asking for, but if it is how much further will the projectile travel while dropping the additional 1.10 m, then just use the two equations again:

you know Vi, A, and D so you can compute T; from T and Vh you an compute the additional distance.