Question

A baseball player throws a baseball with the same initial velocity 5 different times. Each time the ball is thrown at a different angle. The angles are: 45o, 15o, 60o, 25o, and 80o. (a) List the angles based on the amount of time in the air, from the shortest time to the longest time. (b) List the angles based on the horizontal range the baseball travels from the shortest distance to the longest distance.

Answer 1

Answer:

(a) The list of angles based on amount of time in the air; 15° < 25° < 45° < 60°, < 80°

(b) The list of angles based on distance = 80° < 15° < 25° < 60° < 45°

Explanation:

(a) The given parameters are;

The angles in which the ball is thrown, θ = 45°, 15°, 60°, 25°, and 80°

The velocity with which the ball is thrown each time = The same velocity

The amount of time the projectile is in the air given as follows;

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

Where;

t = Half the amount of time the projectile is in the air

θ = The angle of flight of the projectile

u = The initial velocity of the projectile = Constant for all angles in which the ball is thrown

g = The acceleration due to gravity = 9.8 m/s² = Constant

Therefore, we have;

When θ = 45°;

$2 \cdot t = \dfrac{2\cdot u \cdot sin (45^ {\circ}) }{g} = \dfrac{2\cdot u \cdot \dfrac{\sqrt{2} }{2} }{g} = \dfrac{\sqrt{2} \cdot u }{g}$

When θ = 15°;

$2 \cdot t = \dfrac{2\cdot u \cdot sin (15^ {\circ}) }{g} \approx \dfrac{0.517638\cdot u }{g}$

When θ = 60°;

$2 \cdot t = \dfrac{2\cdot u \cdot sin (60^ {\circ}) }{g} = \dfrac{2\cdot u \cdot \dfrac{\sqrt{3} }{2} }{g} = \dfrac{\sqrt{3} \cdot u }{g}$

When θ = 25°;

$2 \cdot t = \dfrac{2\cdot u \cdot sin (25^ {\circ}) }{g} = \dfrac{0.8452365\cdot u }{g}$

When θ = 80°;

$2 \cdot t = \dfrac{2\cdot u \cdot sin (80^ {\circ}) }{g} = \dfrac{1.9696155\cdot u }{g}$

Therefore, the list of the angles based on the amount of time in the air from the shortest time to the longest time is given as follows;

List of angles based on amount of time in the air; 15° < 25° < 45° < 60°, < 80°

(b) The horizontal range, R is given as follows;

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

When θ = 45°

$R = \dfrac{u^2 \cdot sin(2 \times 45 ^{\circ}) }{g} = \dfrac{u^2 \cdot sin(90 ^{\circ}) }{g} = \dfrac{u^2 }{g}$

When θ = 15°

$R = \dfrac{u^2 \cdot sin(2 \times 15 ^{\circ}) }{g} = \dfrac{u^2 \cdot sin(30 ^{\circ}) }{g} = \dfrac{1}{2} \cdot \dfrac{u^2 }{g} = 0.5 \cdot \dfrac{u^2 }{g}$

When θ = 60°;

$R = \dfrac{u^2 \cdot sin(2 \times 60 ^{\circ}) }{g} = \dfrac{u^2 \cdot sin(120 ^{\circ}) }{g} = \dfrac{u^2 }{g} \cdot \dfrac{\sqrt{3} }{2} =0.866025 \cdot \dfrac{u^2 }{g}$

When θ = 25°;

$R = \dfrac{u^2 \cdot sin(2 \times 25 ^{\circ}) }{g} = \dfrac{u^2 \cdot sin(50 ^{\circ}) }{g} = \dfrac{0.7660444 \cdot u^2 }{g}$

When θ = 80°

$R = \dfrac{u^2 \cdot sin(2 \times 80 ^{\circ}) }{g} = \dfrac{u^2 \cdot sin(160 ^{\circ}) }{g} = \dfrac{0.34202 \cdot u^2 }{g}$

Therefore the list of angles based on the horizontal range the baseball travels from the shortest distance to the longest distance is given as follows;

List of angles based on distance = 80° < 15° < 25° < 60° < 45°.