Take the derivative of the function.
By plugging a value into the derivative, you can find the instantaneous velocity.
By setting the derivative equal to zero and solving, you can find the maximums and/or minimums.
Example:
Find the instantaneous velocity at x = 3 and find the maximum height.
f(x) = -x2 + 4
f'(x) = -2x
f'(3) = -2*3 = -6
So the instantaneous velocity is -6.
0 = -2x
0 = x
So the maximum height occurs at x = 0
f(0) = -02 + 4 = 4
So the maximum height is 4.
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It is the gradient (slope) of the line.
1/2mv^2 = mgh
the final velocity assuming that the mass is falling and that air resistance can be ignored but it is acceleration not mass that is important (can be gravity) final velocity is = ( (starting velocity)2 x 2 x acceleration x height )0.5
A stone is thrown with an angle of 530 to the horizontal with an initial velocity of 20 m/s, assume g=10 m/s2. Calculate: a) The time it will stay in the air? b) How far will the stone travel before it hits the ground (the range)? c) What will be the maximum height the stone will reach?
Get the value of initial velocity. Get the angle of projection. Break initial velocity into components along x and y axis. Apply the equation of motion .