Question

A cannonball is fired with an initial velocity of 51m/s at an angle of 35° above the ground.

Answer 1

Answer:

1. $\displaystyle(v_i)_y \approx 29.25\ m/s$
2. $\displaystyle(v_i)_x \approx 41.78\ m/s$  
3. $t\approx 5.97 \ \text{s}$
4. $\triangle x \approx 304.47 \ \text{m}$

Explanation:

Find the vertical component:

• $\displaystyle (v_i)_y = v_i \cdot sin(\theta)$
• $\displaystyle (v_i)_y = 51 \cdot sin(35)$
• $\displaystyle (v_i)_y \approx 29.25$

Find the horizontal component:

• $\displaystyle (v_i)_x = v_i \cdot cos(\theta)$  
• $\displaystyle (v_i)_x = 51 \cdot cos(35)$
• $\displaystyle (v_i)_x \approx 41.78$

We can use this constant acceleration kinematic equation to find the time of flight:

• $v_f=v_i+at$  

Since we don't know the final velocity of the projectile at the end of its trajectory, we can use our knowledge that the final velocity when the projectile reaches its maximum height = 0 m/s.

Therefore, we will solve for half the total time t and double this at the end.

We are going to use the equation in terms of the y-direction since we know the acceleration in the y-direction = g.

• $0=51\cdot sin(35) + (-9.8)t$

Subtract the y-component of the initial velocity from both sides.

• $-51\cdot sin(35) = -9.8t$

Divide both sides by -9.8 to solve for t.

• $t=2.984938597$

Let's double this time to find the total time of flight.

• $2t \approx 5.97$

The time of flight is approximately 5.97 seconds.

Now we can use this kinematic equation to find the displacement in the x-direction:

• $\displaystyle \triangle x = v_i t + \frac{1}{2}at^2$

Using our knowledge, we can assume that the acceleration in the x-direction is 0 m/s² since the question does not specify air resistance.

• $\displaystyle \triangle x=(51)(5.97) + \frac{1}{2} (0)(5.97)^2$
• $\triangle x = (51)(5.97)$
• $\triangle x = 304.47$  

The displacement in the x-direction is approximately 304.47 meters.