help! please..?

Question

A brick falls freely from a high scaffold and hits the ground 3.5 seconds. How high is the scaffold? What is the velocity of the brick just before hitting the ground?

please give the step by step solution with formula..thank you!

Answer 1

1) s = u t + 1/2 a t^2

s = 0 + 1/2 * 9.8 * 3.5^2

s = 60.025 metres high

2) v = u + at
v = 0 + 9.8 * 3.5
v = 34.3 metres per second when it hits the floor.

Answer 2

The brick starts from the scaffold from rest (with zero initial velocity) and reaches the ground after 3.5 seconds. We know that it falls with an acceleration "g", the acceleration due to gravity, that acts downwards and has a value of 9.8m/sec/sec.

The displacement of a body is given by S=u*t + (1/2)*f*t^2, where u is the initial velocity and f is the acceleration in the direction of u.

Here u=0, f=g
So S=(1/2)gt^2 = (1/2)*9.8*(3.5)^2 = 60.025 meter

The final velocity, v, with which the brick hits the ground is given by
v=u+gt. Here u=0. So v=gt = 9.8*3.5 = 34.3 m/sec.

Answer 3

In this problem two values we have to calculate. One is the distance travelled by the brick. and the velocity of the brick at the time of hitting the ground. The known facts are :
The initial velocity of the brick "u" = 0 m/s
The time taken by the brick "t" = 3.5 sec
Acceleration due to gravity "g" = 9.8 m/Sec²
The final velocity "v" = u + gt = 0 + 9.8 x 3.5
.................................................= 34.3 m/s
================================
Again, v² - u² = 2as= 2gh
............(34.3)² - 0² = 2 x 9.8 x h
...........h = (34.3 x 34.3) / (2 x 9.8)
..............= 60.025 metres
===================

Answer 4

I suppose that g= 9.81m/s^-2
the equation for the distance without initial speed

is x= 1/2 (gt^2)

here x= h = 1/2 *9.81 *(3.5)^2 =60.08m

the speed v= g*t = 9.81*3.5 =34.335m/s

Answer 5

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