Beginner mechanics

We know that 2gh= $v^2$

By substituting the given values, we get

$2 \times g\times h$ = $90^{2}$

$\Rightarrow$ h= $\frac {8100}{2 \times 10}$

$\Rightarrow$ h=405 m

Therefore the height of the tower is $\boxed{405 m}$

It can be solved by applying conservation of energy concept. Total energy at the top of the tower = Total energy at the ground We know that : [ total energy = potential energy(PE) + kinetic energy(KE) ]

KE at the top of the tower is zero and so is the PE at the ground. So the equation reduces to :

PE at the top of the tower = KE at the ground

mgh = 1/2 m v^2

gh = 0.5*v^2

h =0.5*v^2/g

h = 0.5*90^2/10

h = 405.

Using the formula $V^2 \ = \ Vo^2 \ + \ 2gh$ we get:

$90^2 \ = \ 0 \ + \ 20 \times h$

$\frac{8100}{20} \ = \frac{20h}{20}$

$h \ = \boxed{405}$

Since it is a free fall acceleration so we will use this equation Vf = Vi + at to get the time first

Vi = 0 as it is a free fall acceleration Then

90 = 10 t

t = 9 seconds

To get the height of the tower which equals the distance covered by the ball we will use this equation

D = Vi t + half A t²

D = 0 + half 10 (9)²

D = 5 × 81 = 405

Given by :

1. $v_{0} = 0$

2. $v_{t} = 90 m/s$

3. $g = 10 m/s^{2}$

And we know that

• $v_{t}^{2} = v_{0}^{2} + 2gh$

So we can substitute the given value to the following equation

$v_{t}^{2} = 0 + 2gh$

$v_{t}^{2} = 2gh$

$h = \frac{v_{t}^{2}}{2g}$

$h = \frac{(90 m/s)^{2}}{2 \times 10 m/s^{2}}$

$h = \frac{8100 m^{2}/s^{2}}{20 m/s^{2}}$

And we get the height of the tower is $\boxed{405 m}$

$90=10t$

$t=9$ seconds

d= $\frac{10*9^2}{2}$

$d=405$ metres

I used the 4th equation of motion. V^2=u^2+2as WHere v is the velocity give, u is the initial velocity which is zero and a is the acceleration given. S is the height.

known : v=0m/s,v'=90m/s,g=10m/s^2 v'^2 = v^2 + 2gh 90^2 = 0^2 + 2 10 h h=405 m

We know that, v^2=2gh By substituting the given values, we get 90^2=2x10xh =>8100=20h =>h=810/2=405m Therefore the height of the tower is 405m

This problem involves an object acting under free-fall. The final velocity of the ball is given the value, $v=90 m/s$ . It is assumed that the initial velocity is zero, $v_{0}=0 m/s$ . Also, since this is an object in free-fall the acceleration considered is the gravitational acceleration $g=9.8 m/s^2\approx(10 m/s^2$ ). The formula used is $v^2=v_{0}^2+2gh$ Solving this equation for $h$ , we get, $h=\frac{v^2-v_{0}^2}{2g}$ . Substituting the given values is as follows, $h=\frac{8100 m^2/s^2-0 m^2/s^2}{20 m/s^2}=405 m$ .

g h = 0.5 v^2

=> h = 0.5 (90^2)/ 10 = 405

GIVEN:

Initial V = 0 m/s

Final V = 90 m/s

G = 10 m/s

SOLUTION:

since V = Initial V + G * T

therefore 90 = 0 + 10T

therefore T = 10 seconds

since S (which is displacement) = Initial V * T + 1/2 G * T^2

therefore S = 1/2 * 10 * 9^2

therefore S = 405m

I'm the only American on here :-\

2gh=v^2 →2 10 h=90^2 →h=405

We know that 2gh=v2 By substituting the given values, we get h=405 m Therefore the height of the tower is 405m

V^2 = Vo^2 +2gh; if Vo = 0,

h = V^2/2g

h= (90)^2/2.10

h = 8100/20 = 405 m.

I prefer to consider this as a graphical method. Consider a velocity-time graph. We wish to find out the distance, i.e. the area under the graph. This is a triangle since acceleration is constant. So we need to find out the maximum velocity (90m/s given in the question) and the time taken. Time taken is v/a = 90/10=9 seconds. So using 1/2 bh, we get 1/2 * 9 * 90 = 405m. So we are done

v=at(the initial velocity is 0 m/s). so t=v/a=90/10=9. again h=(1/2)at^2=(1/2)(10)(9^2)=405 m.

2gh=v^2 Therefore,h=v^2/2g Susbstituting the values, Ans=405

V final square is v initial square + 2ad

90 square = 0 square + 20d

Solve for d would be 405