Question

1. a cannon elevated at an angle of 35° to the horizontal fires a cannonball, which travels the path shown in the diagram below. neglect air resistance and assume the ball lands at the same height above the ground from which it was launched.

(diagram: a cannon at an angle of 35° to the horizontal, with an initial velocity vector v₁, and a parabolic trajectory ending at a bush.)

a. if the ball lands 7.0 × 10² meters from the cannon 10. seconds after it was fired, what is the horizontal component of its initial velocity?
a) 70. m/s
b) 49 m/s
c) 35 m/s
d) 7.0 m/s

b. if the ball’s time of flight is 10. seconds, what is the vertical component of its initial velocity?
a) 9.8 m/s
b) 49 m/s
c) 70. m/s
d) 98 m/s

c. if the angle of elevation of the cannon is decreased from 35° to 30°, the vertical component of the ball’s initial velocity will
a) decrease and its horizontal component will decrease
b) decrease and its horizontal component will increase
c) increase and its horizontal component will decrease
d) increase and its horizontal component will increase

Explanation:

Part a

Step1: Use horizontal motion formula

In horizontal direction (no acceleration, \(a = 0\)), the formula is \(x=v_{x}t\), where \(x\) is the horizontal displacement, \(v_{x}\) is the horizontal component of velocity and \(t\) is the time of flight.
We need to solve for \(v_{x}\), so \(v_{x}=\frac{x}{t}\).
Given \(x = 7.0\times10^{2}\text{ m}\) and \(t = 10\text{ s}\).

Step2: Calculate \(v_{x}\)

Substitute the values into the formula: \(v_{x}=\frac{7.0\times 10^{2}\text{ m}}{10\text{ s}}=70\text{ m/s}\)

Step1: Use vertical motion formula

For vertical motion, the time of flight \(t=\frac{2v_{y}}{g}\) (symmetric projectile motion, \(y - y_{0}=0\), \(v = v_{0}+at\), at the highest - point \(v = 0\) and total time \(t\) from launch to landing).
We need to solve for \(v_{y}\), so \(v_{y}=\frac{gt}{2}\).
Given \(g = 9.8\text{ m/s}^{2}\) and \(t = 10\text{ s}\).

Step2: Calculate \(v_{y}\)

Substitute the values: \(v_{y}=\frac{9.8\text{ m/s}^{2}\times10\text{ s}}{2}=49\text{ m/s}\)

Step1: Recall the component formulas

The initial velocity components are \(v_{x}=v_{0}\cos\theta\) and \(v_{y}=v_{0}\sin\theta\) (assuming initial speed \(v_{0}\) is constant).
When \(\theta\) decreases from \(35^{\circ}\) to \(30^{\circ}\):

• For the vertical component \(v_{y}=v_{0}\sin\theta\), since \(\sin\theta\) is a function that decreases as \(\theta\) decreases in the range \(0^{\circ}<\theta < 90^{\circ}\) (\(\sin35^{\circ}\approx0.574\), \(\sin30^{\circ}=0.5\)), \(v_{y}\) will decrease.
• For the horizontal component \(v_{x}=v_{0}\cos\theta\), since \(\cos\theta\) is a function that increases as \(\theta\) decreases in the range \(0^{\circ}<\theta < 90^{\circ}\) (\(\cos35^{\circ}\approx0.819\), \(\cos30^{\circ}\approx0.866\)), \(v_{x}\) will increase.

Answer:

A. \(70\text{ m/s}\)

Part b