Question

solve it .......................................

Answer 1

Question:-

• A ball is thrown vertically upwards from the top of a tower at 4.9 m/s . it strikes the ground near the base of the tower after 3 seconds . The height of the tower is

Given:-

• Initial velocity= -4.9m/s

• Time = 3 sec

• Acceleration = 9.8 (due to gravity)

Find:-

• Height of the tower-?

Formula to be used:-

• $\boxed{\bf{s= ut+\dfrac{1}{2} at^2}}$

where,

s= distance

u= initial speed

t=time

a=acceleration

Solution:-

Put the given values in the formula we get,

$\tt{s= -4.9\times 3 +\dfrac{1}{2} \times 9.8 \times 3^3}$

$\sf{s= -4.9\times 3 +\dfrac{1}{2} \times 9.8 \times 9}$

$\tt {s= -4.9\times 3 +4.9 \times 9}$

s= 29.4m

Therefore,Height of the tower is 29.4m.

Answer 2

Question :-

A ball is thrown vertically upward from the top of the tower at 4.9 metre per second is strike the ground near the base of the tower after 3 seconds find the height of the tower ?

Solution :-

Given :-

1. initial velocity is -4.9 metre per second
2. time is 3 second
3. acceleration due to gravity is 9.8

what we are going to find here ?

we are going to find here the height of the tower .

Answer :-

to find the height of the tower formula to be used is ;

$s = ut + \frac{1}{2} {at}^{2}$

so,

now putting the values in their sweet table place we can find the answer ;

$s = - 4.93 + \frac{1}{2} \times 9.8 \times {3}^{3}$

$= > s = - 4.9 \times 3 + \frac{1}{2} \times 9.8 \times 9$

$= > s = 29.4m$

hence,

the height of the Tower is 29.4 m .