A projectile is fired with initial speed $$ 500 \mathrm{~m} / \mathrm{s} $$ at an angle 35 degrees above the horizontal over flat ground. What are its maximum height and range?

Question

UpStudy AI · Accepted Answer

We begin by resolving the initial velocity $$ v_0 = 500\,\mathrm{m/s} $$ into its horizontal and vertical components. The launch angle is $$ \theta = 35^\circ $$.

1. The vertical component is
   $$
   v_{0y} = v_0 \sin\theta = 500 \sin 35^\circ,
   $$
   and the horizontal component is
   $$
   v_{0x} = v_0 \cos\theta = 500 \cos 35^\circ.
   $$

2. The maximum height $$ H $$ is reached when the vertical velocity becomes zero. Using the formula
   $$
   H = \frac{v_{0y}^2}{2g},
   $$
   where the acceleration due to gravity is $$ g = 9.8\,\mathrm{m/s^2} $$, we substitute:
   $$
   H = \frac{(500 \sin 35^\circ)^2}{2 \times 9.8}.
   $$
   Simplify the expression:
   $$
   H = \frac{500^2 \sin^2 35^\circ}{19.6}.
   $$
   
   Evaluating numerically:
   - First, compute $$ 500^2 = 250\,000 $$.
   - Using $$ \sin 35^\circ \approx 0.5736 $$, we get
     $$
     \sin^2 35^\circ \approx 0.5736^2 \approx 0.329.
     $$
   - Thus,
     $$
     H \approx \frac{250\,000 \times 0.329}{19.6} \approx \frac{82\,250}{19.6} \approx 4194\,\mathrm{m}.
     $$

3. The range $$ R $$ of the projectile is given by
   $$
   R = \frac{v_0^2 \sin(2\theta)}{g}.
   $$
   Substitute the known values:
   $$
   R = \frac{250\,000 \sin 70^\circ}{9.8}.
   $$
   With $$ \sin 70^\circ \approx 0.9397 $$:
   $$
   R \approx \frac{250\,000 \times 0.9397}{9.8} \approx \frac{234\,925}{9.8} \approx 23\,930\,\mathrm{m}.
   $$

Thus, the maximum height is approximately $$ 4194\,\mathrm{m} $$ and the range is approximately $$ 23\,930\,\mathrm{m} $$.
The projectile reaches a maximum height of about 4,194 meters and travels a range of approximately 23,930 meters.

A projectile is fired with initial speed \( 500 \mathrm{~m} / \mathrm{s} \) at an angle 35 degrees above the horizontal over flat ground. What are its maximum height and range?

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Physics Questions