Please help me to solve these probs today itself..

Question

1)A ball is thrown vertically upwards with a velocity of 20m/s from the top of a multistoried building.The height of the point from where the ball is thrown is 25m from the ground. Find:

a) How high will the ball rise?

b) How long will it be in air before it hits the ground?

2)The position of an object moving along the X-axis is given by x=a bt2 ,where a=8.5 m, b=2.5m/s2 and t is measured in seconds. What is the velocity at t=0s and t=2s? What is the average velocity between t=2s and t=4s?

3) In a 100m race which a sprinter clears in 11s.Calculate his initial uniform acceleration and its duration, if its speed remains constant at 10m/s thereafter?

Answer 1

a 25m

b 3secs

dunno the rest

Answer 2

since the first two have been answered (use kinematics if you want to do it yourself, for 1a) set initial velocity u =20, final v=0 and set

a=-g=-9.81 then use v^2-u^2 = 2ay where y is height, solve for y

for b), use v=u+at and solve for t.

2) x = a bt2 ....i dont know how youve notated this, its not clear, but anyway differentiate the formula you are given (velocity is equal to time derivative of position)

for example, say your formula is this

x = a+bt^2 (t squared)

v = dx/dt = d/dx(a+bt^2) = 2bt

v(x) = 2bt

so at t=0, v=0

at t=2, v=10m/s (this is just for my formula, if yours is different, just redo it, basic differentiation with respect to time to get velocity as a function of position.

Average velocity between two points work out the velocity at each of the points, and add together, then divide by two.

3) x = 100m, t = 11s, v = 10m/s (constant after finish)

constant acceleration

initial velocity u = 0 (assume) final velocity v = 10m/s

uniform acceleration x = ut+1/2at^2

x = 1/2at^2

a=2x/t^2 = 200/11^2 = 1.65 m/s^2 approx

Its duration is 11s because im assuming its uniform and exists until the finish line which his speed remains constant,