Question

4. A golf ball leaves the ground at an angle 0 and hits a tree while moving horizontally at height

h above the ground. If the tree is a horizontal distance of b from the point of projection,what’s the initial velocity interms of b and h

Answer 1

To solve this problem, we can use the equations of motion for projectile motion. We know that the horizontal component of velocity remains constant throughout the motion, while the vertical component changes due to gravity. Let's consider the horizontal and vertical components separately.

Horizontal Component:

The horizontal component of velocity (Vx) remains constant throughout the motion. Therefore, we can write:

Vx = b/t

where t is the time taken by the golf ball to hit the tree.

Vertical Component:

The vertical component of velocity (Vy) changes due to gravity. We can use the following equation to find Vy at any time t:

Vy = u*sin(0) - gt

where u is the initial velocity, 0 is the angle of projection, g is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken by the golf ball to hit the tree.

At the maximum height (h) reached by the golf ball, Vy becomes zero. Therefore, we can write:

h = u^2*sin^2(0)/2g

Solving for u, we get:

u = sqrt(2gh)/sin(0)

Now, we need to find the time taken by the golf ball to hit the tree. We can use the following equation:

b = Vx*t

Solving for t, we get:

t = b/Vx

Substituting this value of t in the equation for Vy, we get:

h = u*sin(0)*t - gt^2/2

Substituting the values of u and t, we get:

h = b*tan(0) - (g*b^2)/(2*Vx^2)

Finally, substituting the value of Vx, we get:

h = b*tan(0) - (g*b^2)/(2*b^2/t^2)

Simplifying, we get:

h = b*tan(0) - (g*t^2)/2

Solving for u, we get:

u = sqrt(2*(h + g*t^2/2)/sin(0))

Therefore, the initial velocity in terms of b and h is given by:

u = sqrt(2*(h + g*(b/Vx)^2/2)/sin(0))