Answer:
3. Height: 16.15 ft; horizontal distance: 160.9 ft.
4. Height: 88.69 ft; horizontal distance: 95.1 ft; time aloft: 4.709 s.
Step-by-step explanation:
You want the maximum height and the horizontal distance traveled for a ball launched at 88 ft/s at an angle of 20° from a height of 2 ft, and for a rocket launched at 78 ft/s at an angle of 75° from ground level. You also want the rocket's time aloft.
Algebra problems like these usually ignore air resistance and model the travel as a quadratic function of time. The height equation is ...
[tex]h(t)=-\dfrac{1}{2}gt^2+v_0t+h_0[/tex]
where v₀ is the initial vertical velocity, and h₀ is the initial height.
When the object is launched at some angle θ with respect to the horizontal, the height h(t) and distance d(t) equations become ...
[tex]h(t)=-\dfrac{1}{2}gt^2+v_0\sin(\theta)t+h_0\\\\d(t)=v_0\cos(\theta)t[/tex]
From the height equation, we can find the time to maximum height as ...
[tex]t_1=\dfrac{v_0\sin(\theta)}{g}[/tex]
And from energy considerations, we can find the time to fall from maximum height as ...
[tex]t_2=\sqrt{\dfrac{2h_0}{g}+t_1^2}[/tex]
The maximum height is then ...
[tex]h_{max}=h_0+\dfrac{(v_0\sin(\theta))^2}{2g}[/tex]
And the horizontal distance traveled is d(t₁ +t₂).
Using the above equations, we find the maximum height to be ...
hmax = 2 + (88·sin(20°))²/(2·32) ≈ 16.15 . . . . feet
The maximum height of the ball is about 16.15 ft.
The time to maximum height is ...
tmax = 88·sin(20°)/32 ≈ 0.940555 . . . . seconds
Then the total time aloft is ...
t = tmax + √(2·2/32 +tmax²) ≈ 1.94537 . . . . seconds
The horizontal distance traveled in this time is ...
d(1.94537) = 88·cos(20°)·1.94537 ≈ 160.9 . . . . feet
The horizontal distance the ball travels is about 160.9 feet.
This problem is solved in the same way. Since the initial height is zero, the total time aloft is double the time it takes to reach the maximum height.
hmax = (78·sin(75°))²/(2·32) ≈ 88.69 . . . . feet
The maximum height of the rocket is about 88.69 ft.
The time aloft is ...
t = 2·78·sin(75°)/32 ≈ 4.709 . . . . seconds
The total time the rocket is in the air is about 4.709 seconds.
The horizontal distance is then ...
d(4.709) = 78·cos(75°)·4.709 ≈ 95.1 . . . . feet
The horizontal distance the rocket travels is about 95.1 feet.
