A projectile is fired with an initial speed of 190 m/s and angle of elevation 60°. The projectile is fired from a position 90 m above the ground. (Recall g ≈ 9.8 m/s^2. Round your answers to the nearest whole number.)
(a) Find the range of the projectile.
b) Find the maximum height reached.

We can solve the projectile motion problem using kinematic equations. Here's a step-by-step approach:

### Given dataWe can solve the projectile motion problem using kinematic equations. Here's a step-by-step approach:

### Given 
- Initial speed: \( v_0 = 190 \, \text{m/s} \)
- Angle of elevation: \( \theta = 60^\circ \)
- Height from which the projectile is fired: \( h_0 = 90 \, \text{m} \)
- Acceleration due to gravity: \( g = 9.8 \, \text{m/s}^2 \)

### Part (a): Finding the Range of the Projectile

To find the range \( R \), we use the equation for the horizontal displacement of a projectile:
\[
R = v_{0x} \cdot t_{\text{total}}
\]
where:
- \( v_{0x} = v_0 \cos \theta \) is the horizontal component of the initial velocity.
- \( t_{\text{total}} \) is the total time of flight.

First, we need to find \( t_{\text{total}} \), the time of flight. The vertical component of the initial velocity is:
\[
v_{0y} = v_0 \sin \theta
\]

The total time of flight \( t_{\text{total}} \) is determined by solving the equation for the vertical displacement. The equation is:
\[
y(t) = h_0 + v_{0y} t - \frac{1}{2} g t^2
\]
Set \( y(t) = 0 \) (since the projectile lands back at ground level), and solve for \( t \).

### Part (b): Finding the Maximum Height Reached

The maximum height \( H \) is reached when the vertical velocity becomes zero. We use the following kinematic equation:
\[
v_y^2 = v_{0y}^2 - 2g(H - h_0)
\]
At maximum height, \( v_y = 0 \), so the equation simplifies to:
\[
0 = v_{0y}^2 - 2g(H - h_0)
\]
Solve for \( H \).

I'll calculate both the range and the maximum height next.### Results:
(a) The range of the projectile is approximately **3,241 meters**.  
(b) The maximum height reached by the projectile is approximately **1,471 meters**.

The total time of flight is about **34 seconds**.

Would you like further details or clarifications?

Here are some related questions for deeper understanding:
1. How does changing the angle affect the range of the projectile?
2. What happens to the maximum height if the initial speed is doubled?
3. How does the initial height (90 m) influence the total time of flight?
4. If the angle were reduced to 45°, what would the new range be?
5. What would the effect be on the range if the projectile were launched from ground level?

**Tip**: The angle of \( 45^\circ \) typically provides the maximum range for a projectile when launched from ground level.

A projectile is fired with an initial speed of 190 m/s and angle of elevation 60°. The projectile is fired from a position 90 m above the ground. (Recall g ≈ 9.8 m/s^2. Round your answers to the nearest whole number.) (a) Find the range of the projectile. (b) Find the maximum height reached.

Solve a projectile motion problem where a projectile is fired at 190 m/s with a 60° angle of elevation from a height of 90 meters. Learn how to calculate the range (3241 meters) and the maximum height (1471 meters) using kinematic equations and trigonometric functions. Suitable for Grades 11-12.

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