Question

A squirrel on a limb near the top of a tree loses its grip on a nut, so that the nut slips away horizontally at a speed of 1.5m/s. If the nut lands at a horizontal distance of 7.2m, how high above the ground is the squirrel?

Answer 1

Answer:

The height of the squirrel  $h =113 \ m$

Explanation:

From the question we are told that  

    The horizontally speed is $v = 1.5 m/s$

    The horizontal  distance $d = 7.2 \ m$

Generally the the time taken is mathematically represented

      $t = \frac{d}{v}$

=>   $t = \frac{7.2}{1.5}$

=>   $t = 4.8 \ s$

Generally from kinematic equation we have that

     $h = ut + \frac{1}{2} gt^2$

Here u  is the initial velocity of nut in the vertical and the value is 0 m/s

     $h = 0 * 4.8 + \frac{1}{2} * 9.8 * (4.8)^2$

=>  $h =113 \ m$