Question

26. Wata, jet comes out from a point on the
ground at a speed v at angle 45° with horizontal
as shown. The speed v of jel so that it can wet
the man upto his head will be (g = 10 m/s)

man 24 m
0345
(1) 10 m/s
(2) 10/2 m/s
(4) 100 mis​

Answer 1

Answer:

400 mis i think sorry if the answer is wrong

Answer 2

Given:

Jet comes from a point at a speed v at an angle 45 with horizontal .

Height of man = 24 m

g = 10 $\frac{m}{s^{2} }$

To Find:

The speed v of jel so that it can wet the man upto his head  

Solution:

Initial direction of the jet = 45 degree with horizontal

Velocity of jet in horizontal direction = v Cos 45.

Velocity of jet in Vertical direction = v Sin 45

The jet will follow projectile motion.

The jet must wet the man from bottom to head

Hence maximum height of the jet = height of man

3rd equation of motion in vertical direction

$v^{2} - u^{2} = 2as\\\\ u = vCos45$

final velocity= 0

a = -g

s=H

substitute all the values to find the velocity of jet;

$(vCos45)^{2} = 2gH\\\\\frac{v^{2} }{2} = 2gH\\ v = \sqrt{(4gH)}\\ v = \sqrt{(4g(24))} \\v= 31 \frac{m}{s}$

Velocity of Jet = 31 meters per seconds.