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.
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 .
To determine the maximum height reached by an object launched with a given initial velocity, you can use the formula for projectile motion. The maximum height is reached when the vertical velocity of the object becomes zero. This can be calculated using the equation: Maximum height (initial velocity squared) / (2 acceleration due to gravity) By plugging in the values of the initial velocity and the acceleration due to gravity (which is approximately 9.81 m/s2 on Earth), you can find the maximum height reached by the object.
To find the instantaneous angular acceleration, you need to know the time rate of change of the instantaneous angular velocity. Without this information, you cannot calculate the instantaneous angular acceleration at t=5.0s.
The maximum height attained by the body can be calculated using the formula: height = (initial velocity)^2 / (2 * acceleration due to gravity). Since the velocity is reduced to half in one second, we can calculate the initial velocity using the fact that the acceleration due to gravity is -9.81 m/s^2. Then, we can plug this initial velocity into the formula to find the maximum height reached.
The instantaneous velocity is the limit of the average velocity, as the time interval tends to zero. If you are not familiar with limits, basically you make the time interval very small and calculate the average velocity.
Multiply the height by 4. The equation to use is h=-16t2 + v0t + h0. Use whatever values you want for v0 and h0, and find the vertex of the parabola. Then double your value of v0, and find the vertex of your new parabola. It will be 4 times as high every time. By the way, to find the vertex, plug in v0/32 for t. Then solve for h.
To find the instantaneous acceleration at t = 45.0s, you need to differentiate the velocity function with respect to time. The acceleration at t = 45.0s is the derivative of the velocity function at that time. Apply the derivative to the velocity function to find the acceleration at t = 45.0s.
To find the instantaneous acceleration of a particle, you would need to know the rate of change of its velocity at that specific moment in time. This can be calculated using calculus by taking the derivative of the velocity function with respect to time. The instantaneous acceleration provides information about how the velocity of the particle is changing at that precise instant.
To find the initial velocity of the kick, you can use the equation for projectile motion. The maximum height reached by the football is related to the initial vertical velocity component. By using trigonometric functions, you can determine the initial vertical velocity component and then calculate the initial velocity of the kick.
To find instantaneous velocity from a position-time graph, you calculate the slope of the tangent line at a specific point on the graph. The slope represents the rate of change of position at that instant, which is equivalent to the velocity at that particular moment.
It depends. If the projectile goes straight up and straight down, its velocity will be zero at the top. If the projectile is a baseball about halfway between the pitcher and the bat, its velocity might be 150 km/h.
To find the maximum height, we first need to separate the initial velocity into its x and y components. Since the initial velocity is given as v = 7.6i + 6.1j, the initial vertical velocity is 6.1 m/s. We can use the kinematic equation for vertical motion: v_f^2 = v_i^2 + 2aΔy, where v_f = 0 at the maximum height. Rearranging the equation to solve for the maximum height, h, we have h = (v_i^2)/2g, where g is the acceleration due to gravity (9.81 m/s^2). Plugging in the values, we find h ≈ 1.88 m.
To find the maximum height, we can use the kinematic equation: ( h = \frac{v^2}{2g} ), where ( h ) is the maximum height, ( v ) is the initial velocity (6 m/s in this case), and ( g ) is the acceleration due to gravity (approximately 9.81 m/s(^2)). Plugging in the values, we get ( h = \frac{6^2}{2*9.81} \approx 1.83 ) meters. Therefore, the stone will reach a maximum height of approximately 1.83 meters.