A golfer is trying to hit their ball over a 20 m tree. The ball leaves their club at an angle of 21° above the horizon. What is the minimum speed the golf ball would need to have after being hit by the club to make it over the tree, in m/s?

To solve this problem, we can use the physics equations of motion. We know that the vertical displacement (height) of the golf ball is 20 m, the initial vertical velocity is what we're trying to find, the acceleration is -9.8 m/s² (due to gravity), and the launch angle is 21°.

The vertical component of the velocity can be found using the equation for the height of a projectile:

h = v_i * t - 0.5 * g * t²

where h is the height, v_i is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.

We can rearrange this equation to solve for the initial vertical velocity:

v_i = (h + 0.5 * g * t²) / t

However, we don't know the time it takes for the golf ball to reach the height of the tree. We can find this using the equation for the time of flight of a projectile launched at an angle:

t = 2 * v_i * sin(θ) / g

where θ is the launch angle.

Substituting this into our equation for the initial vertical velocity gives:

v_i = (h + 0.5 * g * (2 * v_i * sin(θ) / g)²) / (2 * v_i * sin(θ) / g)

Solving this equation for v_i gives the minimum speed the golf ball would need to have after being hit by the club to make it over the tree.

However, this is the vertical component of the velocity. The actual speed of the golf ball is the resultant of the vertical and horizontal components of the velocity. The horizontal component of the velocity can be found using the equation:

v_x = v_i * cos(θ)

The resultant velocity (speed of the golf ball) can then be found using the Pythagorean theorem:

v = sqrt(v_i² + v_x²)

This will give the minimum speed the golf ball would need to have after being hit by the club to make it over the tree, in m/s.