Question

Two stones are projected horizontally from the top of a tower of height 78.4 m with velocities 25 ms
and 45 ms. The separation between them on reaching the ground if both are projected in the same
direction is
a) 40 m
b) 80 m
c) 120 m
d) 160 m
​

Answer 1

Explanation:

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Answer 2

(b)The separation between them on reaching the ground is 80 m.

Explanation:

The height of the tower, h = 78 .4 m

The velocity of the first stone $v_1 = 25 m/s$ ,  and second stone is  $v_2 = 45 m/s$.

Now use equation of motion,

$h = ut+ \frac{1}{2}gt^2$  

Here, h is the height of the tower and u is the initial velocity.

Height is same for both the stone and initial velocity is zero for both.

Therefore,

$78.4m=0+\frac{1}{2}\times(9.8m/s^2)t^2$

t = 4 s.

Range for the first stone,

$d_1 = v_1\times t$

$d_1=25m/s\times4s$

$d_1=100m$

Range for the second stone,

$d_2 = v_2\times t$

$d_2=45m/s\times4s$

$d_2=180m$

Now separation between them

= 180 m - 100 m

= 80 m.

Thus, the separation between them on reaching the ground is 80 m.

#Learn More: equation of motion.

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