Question

A baseball is thrown at 25 m/sec at an angle of 30 degrees above the horizontal.
a) How far away does the ball land?
b) What is the minimum height reached?

- A football is kicked at an angle 45 degrees above the horizontal. What is its initial speed if it covers 150 ft?

-A golf ball leaves a tee at 60 m/sec. and at an angle of 50 degrees above the horizontal. Find:
a) The total time of flight
b) The maximum height reached
c) The horizontal distance covered.

Answer 1

Task:

1. A baseball is thrown at 25 m/sec at an angle of 30 degrees above the horizontal.

a) How far away does the ball land?

b) What is the maximum height reached?

2. A football is kicked at an angle 45 degrees above the horizontal. What is its initial speed if it covers 150 ft?

3. A golf ball leaves a tee at 60 m/sec. and at an angle of 50 degrees above the horizontal. Find:

a) The total time of flight

b) The maximum height reached

c) The horizontal distance covered.

Solution:

1.

$$v _ {0} = 2 5 \frac {m}{s}$$ $$\alpha = 3 0 {}^ {\circ}$$ $$R = v _ {0} \cos \alpha \cdot t$$ $$H = v _ {0} \sin \alpha \cdot \frac {t}{2} - g \cdot \frac {\left(\frac {t}{2}\right) ^ {2}}{2} = \frac {v _ {0} ^ {2} \sin^ {2} \alpha}{2 g}$$ $$t = \frac {R}{v _ {0} \cos \alpha}$$ $$v _ {0} \sin \alpha \cdot \frac {\left(\frac {R}{v _ {0} \cos \alpha}\right)}{2} - g \cdot \frac {\left(\frac {R}{\frac {v _ {0} \cos \alpha}{2}}\right) ^ {2}}{2} = \frac {v _ {0} ^ {2} \sin^ {2} \alpha}{2 g}$$ $$R = 55.175 \, \text{m}$$ $$H = \frac {v _ {0} ^ {2} \sin^ {2} \alpha}{2 g} = 7.964 \, \text{m}$$

2.

$$\alpha = 45{}^{\circ}$$ $$R = 150 \, \text{ft}$$ $$R = v _ {0} \cos \alpha \cdot t$$ $$H = v _ {0} \sin \alpha \cdot \frac {t}{2} - g \cdot \frac {\left(\frac {t}{2}\right) ^ {2}}{2} = \frac {v _ {0} ^ {2} \sin^ {2} \alpha}{2 g}$$ $$t = \frac {R}{v _ {0} \cos \alpha}$$ $$v _ {0} \sin \alpha \cdot \frac {\left(\frac {R}{v _ {0} \cos \alpha}\right)}{2} - g \cdot \frac {\left(\frac {R}{\frac {v _ {0} \cos \alpha}{2}}\right) ^ {2}}{2} = \frac {v _ {0} ^ {2} \sin^ {2} \alpha}{2 g}$$ $$v _ {0} = 69.444 \, \text{ft/s}$$

3.

$$v _ {0} = 60 \, \text{m/s}$$ $$\alpha = 50{}^{\circ}$$ $$R = v _ {0} \cos \alpha \cdot t$$ $$H = v _ {0} \sin \alpha \cdot \frac {t}{2} - g \cdot \frac {\left(\frac {t}{2}\right) ^ {2}}{2} = \frac {v _ {0} ^ {2} \sin^ {2} \alpha}{2 g}$$ $$t = \frac {R}{v _ {0} \cos \alpha}$$ $$v _ {0} \sin \alpha \cdot \frac {\left(\frac {R}{v _ {0} \cos \alpha}\right)}{2} - g \cdot \frac {\left(\frac {\frac {R}{v _ {0} \cos \alpha}}{2}\right) ^ {2}}{2} = \frac {v _ {0} ^ {2} \sin^ {2} \alpha}{2 g}$$ $$R = 361.397 \, \text{m}$$ $$H = \frac {v _ {0} ^ {2} \sin^ {2} \alpha}{2 g} = 107.674 \, \text{m}$$ $$t = \frac {R}{v _ {0} \cos \alpha} = 9.371 \, \text{s}$$

**Answer:**

1. a) $R = 55.175 \, \text{m}$

b) $H = \frac{v_0^2 \sin^2 \alpha}{2g} = 7.964 \, \text{m}$

2. $v_{0} = 69.444 \frac{ft}{s}$

3. a) $t = \frac{R}{v_0 \cos \alpha} = 9.371 \, \text{s}$

b) $H = \frac{v_0^2 \sin^2 \alpha}{2g} = 107.674 \, \text{m}$

c) $R = 361.397 \, \text{m}$