from the top of a tower t ball is thrown vertically upward when the ball reaches distance h below t it's speed was double if what it was at height h above t show that greatest height attained by it above t is 5/2h
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To solve this problem, we can use the equations of motion for vertical motion. Let's denote the height above the tower as "h1" when the ball is thrown upward and "h2" when the ball reaches the distance "h" below the tower.
Let's calculate the initial velocity of the ball, denoted as "v0," when it was thrown from the top of the tower. At the highest point of its trajectory, the velocity of the ball becomes zero.
Using the equation of motion for vertical motion:
v = u + at,
where:
v = final velocity (0 m/s at the highest point),
u = initial velocity, and
a = acceleration due to gravity (-9.8 m/s^2).
We can rearrange the equation to find the initial velocity:
u = v - at.
Substituting the variables:
0 = v0 - (9.8)t,
v0 = 9.8t.
Now, let's consider the speed of the ball at distance "h" below the top of the tower. The speed is double the initial speed at height h above the top of the tower.
Using the equation of motion:
v^2 = u^2 + 2as,
where:
v = final velocity,
u = initial velocity,
a = acceleration due to gravity (-9.8 m/s^2), and
s = displacement.
Rearranging the equation, we have:
v = √(u^2 + 2as).
Substituting the variables:
v = √((9.8t)^2 + 2(-9.8)(h)).
Since the speed at height "h" below the top is double the initial speed at height "h" above the top, we can write:
2v0 = √((9.8t)^2 + 2(-9.8)(h)).
Substituting v0 = 9.8t, we have:
2(9.8t) = √((9.8t)^2 + 2(-9.8)(h)).
Squaring both sides of the equation, we get:
4*(9.8t)^2 = (9.8t)^2 + 2*(-9.8)*(h).
Simplifying the equation, we have:
4 * 9.8^2 * t^2 = 9.8^2 * t^2 - 19.6 * h.
Dividing both sides by 9.8^2 * t^2, we get:
4 = 1 - 2h / (9.8t^2).
Rearranging the equation, we have:
2h = 4 * 9.8 * t^2 - 9.8 * t^2,
2h = 39.2 * t^2 - 9.8 * t^2,
2h = 29.4 * t^2,
h = 14.7 * t^2.
Now, we can find the greatest height attained by the ball above the top, denoted as "H."
Since h1 is the initial height above the top and h2 is the distance below the top where the speed becomes double, we have:
H = h1 + h2,
H = h + h,
H = 2h.
Substituting the value of h from the previous equation, we get:
H = 2 * 14.7 * t^2,
H = 29.4 * t^2.
Comparing this equation with the given equation H = (5/2)h, we see that:
H = 29.4 * t^2 = (5/2) * 14.7 * t^2,
H = 14.7 * t^2.
Therefore, the greatest height attained by the ball above the top, H, is equal to 5/2 times the distance h. Hence, the greatest height attained by the ball above the top is (5/2)h.