Question

A stone is dropped from top of the tower 400m and at the same time another stone is projected upward vertically from the ground with velocity of 100m/s find where and when the two stones will meet

Answer 1

Answer:

x=320m

Explanation:

case 1

1g=10m/s

let we assume that the stones meet at 'x' m high above the ground and 400-x m from the tower.

1u=0m/s

400-x=1/2×10×t2

x=400-5t2.......eq1

case2

u=100m/s

a=-g=-10m/s2

u=100t+1/2 (-10) t2

=100t-5t

x=100t-5t2..........eq2

eq1 an 2

400-5t2=100-5t2

t=4s........eq3

substitute eq3 in 1

x=400-5t2

=400-5 (16)

=400-80

=320m

[x=320m]

Answer 2

Answer:

The stones meet each other at 78.4m below the top of the tower after time 4 seconds.

Explanation:

Consider that two stones meet each other at distance 'h' from the top of the tower after time 't'.

We have given, the height of the tower = 400m

Consider the downward motion of first stone:

$h=ut+\frac{1}{2}gt^2$

where, u=0 and g = 9.8ms⁻²,

$h=\frac{1}{2}gt^2$

$h=\frac{9.8}{2}t^2$                                                                          ...............(1)

The velocity of stone projected upward from the ground = 100m/s

Let's consider the upward motion of the second stone:

$400-h=100t-\frac{1}{2} *(9.8)t^2$                                          ..................(2)

From the eq. (1) and (2) , we get;

$400-\frac{1}{2}*9.8*t^2=100t-\frac{1}{2} *9.8*t^2$

$400=100t$

$t=4s$

Substitute the value of time t = 4s in the equation (1);

$h=\frac{1}{2}*(9.8)*(4)^2$

$h=\frac{1}{2}*9.8*16$

$h=78.4m$

Therefore, after 4sec both stones meet each other at distance 78.4m from the top of the tower.