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Let's solve this problem step by step. Here's the problem restated:

A golfer hits a ball from ground level with an initial velocity of \( 21 \, \text{m/s} \) at an angle of \( 60^\circ \) above the horizontal. We are to find:
1. Maximum height of the ball.
2. Range of the ball.
3. Speed of the ball at its maximum height.

### Known Data:
- Initial velocity, \( u = 21 \, \text{m/s} \)
- Angle of projection, \( \theta = 60^\circ \)
- Gravitational acceleration, \( g = 9.81 \, \text{m/s}^2 \)

### Part (a): Maximum Height of the Ball
The maximum height is achieved when the vertical velocity becomes zero. We can use the following kinematic equation for vertical motion:

\[
v_y^2 = u_y^2 - 2 g h
\]
At maximum height, the final vertical velocity \( v_y = 0 \), so:

\[
0 = u_y^2 - 2 g h \implies h = \frac{u_y^2}{2g}
\]
The vertical component of the initial velocity is:

\[
u_y = u \sin \theta = 21 \sin 60^\circ
\]

Now, we can calculate \( h \).

### Part (b): Range of the Ball
The range (horizontal distance) is the total horizontal displacement when the ball returns to the same level. The range is given by the equation:

\[
R = \frac{u^2 \sin 2\theta}{g}
\]
We can plug in the known values to calculate \( R \).

### Part (c): Speed at Maximum Height
At maximum height, the vertical velocity is zero, so the speed of the ball is just its horizontal component. The horizontal component of velocity remains constant throughout the flight:

\[
v_x = u \cos \theta = 21 \cos 60^\circ
\]
This will give us the speed at the maximum height.

Let me calculate these values now.Here are the results for the three parts:

### Part (a): Maximum Height of the Ball
The maximum height reached by the ball is approximately:
\[
h_{\text{max}} = 16.86 \, \text{m}
\]

### Part (b): Range of the Ball
The range (horizontal distance) covered by the ball is approximately:
\[
R = 38.93 \, \text{m}
\]

### Part (c): Speed at Maximum Height
The speed of the ball at its maximum height (which is the horizontal component of the velocity) is:
\[
v_{\text{max height}} = 10.5 \, \text{m/s}
\]

Would you like further details on any part or have questions?

Here are 5 related questions for deeper understanding:
1. How would the maximum height change if the angle of projection were smaller?
2. What is the time of flight of the ball in this situation?
3. How does the range change if air resistance is considered?
4. What is the significance of the horizontal and vertical components of velocity?
5. How would the initial speed affect the range if the angle remains constant?

**Tip:** For projectile motion, the time to reach maximum height is half of the total time of flight.

A golfer hits a ball, from ground level on a horizontal surface. The initial velocity of the ball is 21 m/s at an angle of 60° above the horizontal. Find the maximum height, range, and speed at maximum height.

Solve this projectile motion problem where a golfer hits a ball at 21 m/s at a 60° angle. Learn to calculate the maximum height (16.86 m), range (38.93 m), and speed at maximum height (10.5 m/s).

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