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A ball is dropped from rest at a height of 2 metres. Assuming acceleration due to gravity (g) is 10m/s^2 calculate the velocity of the ball just before it hits the floor.

Using the equation V 2 = U 2 + 2 a s, where V = final velocity, U = initial velocity, a = acceleration, s = displacement, we can substitute in the values given in the question to calculate the final velocity...
PB
Answered by Phil B. • Physics tutor
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Difference between Qui/Que?

Qui and Que are relative pronouns. That means that they are used to form one sentence out of two different ones to avoid repetition. Although both pronouns serve the same purpose, they are used for different...
EL
Answered by Eoghan L. • French tutor
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When a particle travels in a circle of radius r, at constant speed v, what is its acceleration

v 2 /r, towards the center of the circle.Remember that acceleration is the rate of change of velocity , not merely of speed. This means that the change in direction is important. In a unit of time, the veloc...
Answered by • Physics tutor
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Find the stationary points on the curve: y = x^3 + 3x^2 +2x+5

Firstly differentiate the function: f(x) = x 3 + 3x 2 + 2x + 5 (function)f'(x) = 3x 2 + 6x + 2 (gradient function) Stationary points are points where the graph has a gradient of zero 3x 2 + 6x + 2 = 0 In ord...
NC
Answered by Nicolas C. • Maths tutor
7443 Views

Work out the gradient of the curve y=x^3(x-3) at the point (3,17)

First simplify the equation of the curve y= x^4 - 3x^3 .The gradient is the differential.To differentiate, bring down the power and take one from it.x^4 becomes 4x^3-3x^3 becomes (-3x3)= -9x^2dy/dx= 4x^3 - 9...
SM
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