Question

Please help!!

Answer 1

Answer:

(a) The time the ball stays in the air is approximately 4.66 seconds

(b) The distance from the cannon at which the ball will hit the ground is approximately 83.18 meters

(c) The maximum altitude of the ball is approximately 26.62 m

Explanation:

The given parameters of the question are;

The initial velocity at which a baseball cannon fires balls, u = 29 m/s

The angle at which the cannon is tilted, θ = 52°

(a) The time duration the ball stays in the air is given by the time of flight of the projected ball as follows;

$2 \cdot t = \dfrac{2 \cdot u \cdot sin (\theta)}{g}$

Where;

t = The time it takes the baseball to reach maximum height

2·t = The total time of flight = The time the ball stays in the air

θ = The angle at which the ball is tilted = 52°

g = The acceleration due to gravity ≈ 9.81 m/s²

u = The initial velocity = 29 m/s

Therefore, we have;

$2 \cdot t = \dfrac{2 \times 29 \ m/s\times sin (52^{\circ})}{9.81 \ m/s^2} \approx 4.66 \ s$

The time the ball stays in the air, 2·t ≈ 4.66 seconds

(b) The distance from the cannon at which the ball will hit the ground is given by the horizontal range, 'R', of the projectile as follows;

$Horizontal \ range, R = \dfrac{u^2 \cdot sin(2 \cdot \theta) }{g}$

∴ The distance from the cannon at which the ball will hit the ground = R

$Horizontal \ range, R = \dfrac{(29 \ (m/s))^2 \cdot sin(2 \times 52^{\circ}) }{9.81 \ m/s^2} \approx 83.18 \, m$

The distance from the cannon at which the ball will hit the ground = R ≈ 83.18 meters

(c) The maximum altitude of the ball is equal to the maximum height reached by the ball, H, which is given as follows;

$H = \dfrac{u^2 \cdot sin^2 (\theta)}{2 \cdot g}$

Therefore, we have;

$H = \dfrac{(29 \ m/s)^2 \times sin^2 (52^{\circ})}{2 \times 9.81 \ m/s^2} \approx 26.62 \ m$

The maximum altitude of the ball ≈ 26.62 m