Question

1. a) find the range of the projectile and b) find the max height it reaches. image shows a circle near a vertical line, with velocity components: ( v_y = 25 , \text{m/s} ) (horizontal dashed arrow), ( v_x = 12 , \text{m/s} ) (vertical dashed arrow), and a solid velocity vector ( v ) from the circle to the tip of the dashed arrows.

Explanation:

Part (a): Range of the Projectile

Step 1: Recall the range formula for projectile motion

The range \( R \) of a projectile is given by \( R = \frac{2v_x v_y}{g} \), where \( v_x \) is the horizontal velocity, \( v_y \) is the vertical velocity, and \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).

Step 2: Identify given values

From the diagram, \( v_x = 12 \, \text{m/s} \) (horizontal component) and \( v_y = 25 \, \text{m/s} \) (vertical component).

Step 3: Substitute values into the formula

Substitute \( v_x = 12 \), \( v_y = 25 \), and \( g = 9.8 \) into \( R = \frac{2v_x v_y}{g} \):
\[
R = \frac{2 \times 12 \times 25}{9.8}
\]

Step 4: Calculate the range

First, compute the numerator: \( 2 \times 12 \times 25 = 600 \).
Then divide by \( 9.8 \): \( R = \frac{600}{9.8} \approx 61.22 \, \text{m} \).

Part (b): Maximum Height of the Projectile

Step 1: Recall the maximum height formula

The maximum height \( H \) of a projectile is given by \( H = \frac{v_y^2}{2g} \), where \( v_y \) is the vertical velocity and \( g = 9.8 \, \text{m/s}^2 \).

Step 2: Substitute values into the formula

Substitute \( v_y = 25 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \) into \( H = \frac{v_y^2}{2g} \):
\[
H = \frac{25^2}{2 \times 9.8}
\]

Step 3: Calculate the maximum height

First, compute \( 25^2 = 625 \).
Then compute the denominator: \( 2 \times 9.8 = 19.6 \).
Finally, divide: \( H = \frac{625}{19.6} \approx 31.89 \, \text{m} \).

Answer:

s:
a) The range of the projectile is approximately \( \boldsymbol{61.22 \, \text{m}} \).
b) The maximum height reached is approximately \( \boldsymbol{31.89 \, \text{m}} \).