a ball is shot at a ground level at an angle of 60 degrees with an initial velocity of 100m/s

a. find the velocity of a ball after 2 seconds

b. maximum height of the ball

c. range

d. time of flight

We will solve the problem using the equations of motion under constant acceleration (assuming no air resistance). The gravitational acceleration \( g \) is taken as \( 9.8 \, \text{m/s}^2 \).

### Given:
- Initial velocity \( u = 100 \, \text{m/s} \)
- Angle of projection \( \theta = 60^\circ \)
- Gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \)

#### Horizontal and Vertical Components of Initial Velocity:
1. Horizontal component \( u_x = u \cos\theta = 100 \cos 60^\circ = 100 \times \frac{1}{2} = 50 \, \text{m/s} \)
2. Vertical component \( u_y = u \sin\theta = 100 \sin 60^\circ = 100 \times \frac{\sqrt{3}}{2} = 86.6 \, \text{m/s} \)

---

### a. Velocity of the ball after 2 seconds

The velocity components after time \( t = 2 \, \text{seconds} \) are calculated as:

1. **Horizontal velocity** (remains constant):
   \[
   v_x = u_x = 50 \, \text{m/s}
   \]

2. **Vertical velocity** (using \( v_y = u_y - g t \)):
   \[
   v_y = 86.6 - 9.8 \times 2 = 86.6 - 19.6 = 67 \, \text{m/s}
   \]

Now, the magnitude of the velocity at \( t = 2 \, \text{seconds} \) is:
\[
v = \sqrt{v_x^2 + v_y^2} = \sqrt{50^2 + 67^2} = \sqrt{2500 + 4489} = \sqrt{6989} \approx 83.6 \, \text{m/s}
\]

---

### b. Maximum Height of the Ball

At maximum height, the vertical velocity becomes zero. We can find the time to reach maximum height using \( v_y = u_y - g t_{max} \), setting \( v_y = 0 \):

\[
0 = 86.6 - 9.8 t_{max}
\]
\[
t_{max} = \frac{86.6}{9.8} \approx 8.84 \, \text{seconds}
\]

The maximum height can be calculated using the equation:
\[
h_{max} = u_y t_{max} - \frac{1}{2} g t_{max}^2
\]
\[
h_{max} = 86.6 \times 8.84 - \frac{1}{2} \times 9.8 \times (8.84)^2
\]
\[
h_{max} = 765.4 - 380.9 = 384.5 \, \text{m}
\]

---

### c. Range of the Ball

The range is the horizontal distance traveled during the entire time of flight. First, we calculate the total time of flight by doubling the time to reach the maximum height (since the motion is symmetric):

\[
T = 2 \times t_{max} = 2 \times 8.84 = 17.68 \, \text{seconds}
\]

The range \( R \) is given by:
\[
R = u_x \times T = 50 \times 17.68 = 884 \, \text{m}
\]

---

### d. Time of Flight

As calculated earlier:
\[
T = 17.68 \, \text{seconds}
\]

---

### Final Answers:
- **(a)** Velocity after 2 seconds: \( \approx 83.6 \, \text{m/s} \)
- **(b)** Maximum height: \( 384.5 \, \text{m} \)
- **(c)** Range: \( 884 \, \text{m} \)
- **(d)** Time of flight: \( 17.68 \, \text{seconds} \)

---

Do you want more details on any part of the solution or have any questions?

Here are 5 related questions to explore further:
1. What would happen to the range if the angle of projection were 45° instead of 60°?
2. How would the maximum height change if the initial velocity were increased to 150 m/s?
3. What would be the horizontal range if the ground was inclined at 10°?
4. How does air resistance affect the time of flight and range in real-life situations?
5. What is the significance of the angle of 45° in projectile motion for achieving maximum range?

**Tip:** The angle that provides the maximum range for a projectile in a vacuum (no air resistance) is always 45°.

A ball is shot at ground level at an angle of 60 degrees with an initial velocity of 100m/s. Find the velocity of the ball after 2 seconds, the maximum height of the ball, the range, and the time of flight.

Solve a projectile motion problem where a ball is shot at 60° with an initial velocity of 100 m/s. Learn to calculate the velocity after 2 seconds, the maximum height, range, and time of flight using kinematic equations and trigonometry. Suitable for Grades 11-12.

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