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 with initial velocity ( v_i ) at 35° to horizontal, trajectory shown as dashed curve, landing at a bush on the same horizontal line.)

a. if the ball lands ( 7.0 \times 10^2 ) 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

handwritten: ( \frac{700m}{10s} = 70 m/s )

Explanation:

Step1: Recall the horizontal motion formula

In projectile motion (with no air - resistance), the horizontal motion is a uniform - motion. The formula for horizontal displacement is \(x = v_{x}t\), where \(x\) is the horizontal displacement, \(v_{x}\) is the horizontal component of the initial velocity, and \(t\) is the time of flight.
We can re - arrange the formula to solve for \(v_{x}\): \(v_{x}=\frac{x}{t}\).

Step2: Substitute the given values

We are given that \(x = 7.0\times10^{2}\text{ m}\) and \(t = 10\text{ s}\).
Substitute these values into the formula \(v_{x}=\frac{x}{t}\), so \(v_{x}=\frac{7.0\times 10^{2}\text{ m}}{10\text{ s}}\).

Answer:

\(v_{x}=70\text{ m/s}\), so the answer is A. 70 m/s.