Answer:
h = 61.25 m
Step-by-step explanation:
It is given that,
The initial velocity of the ball, v = 60 m/s
It is thrown from a height of 5 feet,
We need to find the maximum height it reaches. The height reached by the projectile as a function of time t is given by :
Putting all the values,
.....(1)
For maximum height, put
Put t = 1.875 in equation (1)
So, the maximum height reached by the ball is 61.25 m.
Hi there!
We can calculate dy/dx using implicit differentiation:
xy + y² = 6
Differentiate both sides. Remember to use the Product Rule for the "xy" term:
(1)y + x(dy/dx) + 2y(dy/dx) = 0
Move y to the opposite side:
x(dy/dx) + 2y(dy/dx) = -y
Factor out dy/dx:
dy/dx(x + 2y) = -y
Divide both sides by x + 2y:
dy/dx = -y/x + 2y
We need both x and y to find dy/dx, so plug in the given value of x into the original equation:
-1(y) + y² = 6
-y + y² = 6
y² - y - 6 = 0
(y - 3)(y + 2) = 0
Thus, y = -2 and 3.
We can calculate dy/dx at each point:
At y = -2: dy/dx = -(-2) / -1+ 2(-2) = -2/5.
At y = 3: dy/dx = -(3) / -1 + 2(3) = -3/5.