Question

a monkey can jump a maximum of horizontal distance of 20m. the velocity of the monkey is?

Answer 1

Let vx be the horizontal velocity and vy be the vertical velocity. 
Now horizontal range L = vx t. Here t is time. In this time moneky goes up and comes down. Time to go up is t/2. Terefore 
vy = g t/2. 
Velocity = v^2 = vx^2 + vy^2=g^2 t^2/4 +L^2/t^2=v^2 
To simpliy we assume vx=vy=a 
a= gt/2 
a = L/t. You have 
L/t = gt/2, t^2= 2L/g= 2*20/9.8=4.08, t=2.02 
a= 40/2.02= 19.8 
velocity = sqrt(2a^2), v= sqrt (vx^2 +vy^2)= 
This gives velocity= 28 m/s

Answer 2

The velocity of the monkey is 14ms-1.

Given:

The monkey can jump a maximum horizontal distance of 20m.

To Find:

The velocity of the monkey.

Solution:

To find the velocity of the monkey we will follow the following steps:

As we know,

The horizontal distance(r) is given by the formula

$r \: = {u}^{2} \frac{sin2x}{g}$

Here, u is the velocity and g is the acceleration constant = 10ms-2.

Given the maximum horizontal distance which is at an angle of x = 45° because then 2x = 90°. so, the sin2x and horizontal distance will be maximum.

sin2(45) = sin90° = 1

Now,

$r = \frac{ {u}^{2} }{g}$

So,

$u² = r × g = 20 × 10 = 200$

$u \: = \sqrt{200} = 14m {s}^{ - 1}$

Henceforth, the velocity of the monkey is 14ms-1.