Question

(a) In the absence of air resistance, a golf ball projected from the point P follows a parabolic path and lands at the point Q. The initial velocity of the ball is u and at an angle of a with the horizontal

(i) Derive an expression for the horizontal distance PQ in terms of u, a, and the acceleration due to gravity g.
(ii) What is the value of the angle of projection a for the distance PQ to be maximum?

(b) A motorcycle stuntman who is moving horizontally takes off from a point 15.0 m above the ground and lands 60.0 m away as shown in the figure. Calculate:
(i) the time between taking off and landing
(ii) the speed of the motorcycle at take-off. ​

Answer 1

a.

i. t = u^2 * sin(a) * cos(a) / g

ii. 90 degree

Step-by-step explanation:

a.(i) The horizontal distance PQ can be expressed as:

PQ = u * cos(a) * t

Where t is the time taken for the ball to reach the ground, which can be determined using the vertical motion equation:

v = u * sin(a) - g * t

where v is the final velocity (0 m/s at the ground) and g is the acceleration due to gravity (9.8 m/s^2). Solving for t, we have:

t = (u * sin(a)) / g

Substituting t back into the expression for PQ:

PQ = u * cos(a) * (u * sin(a)) / g

t = u^2 * sin(a) * cos(a) / g

(ii) To find the angle of projection for maximum PQ, we need to find the maximum of PQ with respect to a. Taking the derivative of PQ with respect to a, setting it equal to zero, and solving for a, we have:

dPQ/da = (u^2 * cos^2(a) - u^2 * sin^2(a)) / g = 0

This gives us:

sin(2a) = 0

The maximum value of PQ occurs at:

a = 90° or a = 270°

So the angle of projection for maximum PQ is 90 degrees.