from the top of a tower P,a ball is thrown vertically upwards.when the ball reaches a distance h below P,its speed is double of what it was at height h above P.show that the greatest height attained by the ball above P is 5h/3.
To solve this problem, we can use classic equations of motion. Let's assume the initial height above P is h1, and the final height below P is h2.
First, let's consider the ball's motion from the starting point above P to the point h below P. The initial velocity (u1) is positive because the ball is thrown upwards, and the final velocity (v1) is negative because the ball is moving downwards. The acceleration (a) due to gravity is always negative. We also know that the speed at the point h below P is double the speed at the initial height h1 above P.
The equation relating initial velocity (u1), final velocity (v1), and acceleration (a) is:
(v1)^2 = (u1)^2 + 2ah
Since the speed at h below P is double the speed at h1 above P, we can write this equation as:
(2v1)^2 = (u1)^2 + 2a(-h1)
4(v1)^2 = (u1)^2 - 2ah1 ---- Equation 1
Now, let's consider the ball's motion from the starting point above P to the highest point it attains above P. At the highest point, the velocity becomes zero.
Using the equation of motion:
(v2)^2 = (u2)^2 + 2a(-h2)
Since the velocity at the highest point is zero, we can write this equation as:
0 = (u2)^2 + 2a(-h2)
(u2)^2 = 2ah2 ---- Equation 2
To find the greatest height attained by the ball above P, we need to find h2 in terms of h. Since the total height is h1 + h2, we can write:
h1 + h2 = h
Rearranging, we have:
h2 = h - h1
Now, let's substitute equations 1 and 2 into the expression for h2:
(u2)^2 = 2ah2
(u2)^2 = 2a(h - h1)
Substituting u1^2 with 4(v1)^2 from equation 1:
4(v1)^2 = 2a(h - h1)
Simplifying:
2(v1)^2 = a(h - h1)
Now, let's substitute v1^2 with (u1)^2 + 2ah from equation 1:
2[(u1)^2 + 2ah1] = a(h - h1)
Expanding:
2(u1)^2 + 4ah1 = ah - ah1
Grouping similar terms:
2(u1)^2 + 5ah1 = ah
Now, let's substitute (u1)^2 with 2ah1 from equation 2:
2(2ah1) + 5ah1 = ah
4ah1 + 5ah1 = ah
9ah1 = ah
Dividing both sides by ah:
9h1 = h
h1 = h/9
Finally, we can substitute h1 into the expression for h2:
h2 = h - h1
h2 = h - h/9
h2 = 8h/9
Therefore, the greatest height attained by the ball above P is h2 which equals 8h/9.
To calculate the maximum height, substitute this value of h2 back into the expression h1 + h2:
h1 + h2 = (h/9) + (8h/9) = 9h/9 = h
So, the final result shows that the greatest height attained by the ball above P is indeed 5h/3.