Question

monkey throws a banana horizontally from the top of a tree. The banana hits the ground 4 seconds later and lands 80 meters from the base of the tree. How fast did the monkey throw the banana?

Answer 1

Given:

A monkey throws a banana horizontally from the top of a tree. The banana hits the ground 4 seconds later and lands 80 meters from the base of the tree.

To find:

Velocity with which banana was thrown?

Calculation:

• This is an example of height-ground projectile.

• Let Velocity of projection be v horizontally from top of tree.

$\rm \: h = u_{y}(t) + \dfrac{1}{2} g {t}^{2}$

$\rm \implies\: h = 0 + \dfrac{1}{2} g {t}^{2}$

$\rm \implies\: h = \dfrac{1}{2} g {t}^{2}$

$\rm \implies\:t = \sqrt{ \dfrac{2h}{g} }$

Now, range will be :

$\rm \: R = u_{x}(t)$

$\rm \implies\: R = v \times t$

$\rm \implies\: R = v \times \sqrt{ \dfrac{2h}{g} }$

$\rm \implies\: 80 = v \times 4$

$\rm \implies\:v = 20 \: m {s}^{ - 1}$

So, Velocity of projection is 20 m/s.

Answer 2

Answer:

$Given:</p><p></p><p>A monkey throws a banana horizontally from the top of a tree. The banana hits the ground 4 seconds later and lands 80 meters from the base of the tree.</p><p></p><p>To find:</p><p></p><p>Velocity with which banana was thrown?</p><p></p><p>Calculation:</p><p></p><p>This is an example of height-ground projectile.</p><p></p><p>Let Velocity of projection be v horizontally from top of tree.</p><p></p><p>\rm \: h = u_{y}(t) + \dfrac{1}{2} g {t}^{2}h=uy(t)+21gt2</p><p></p><p>\rm \implies\: h = 0 + \dfrac{1}{2} g {t}^{2}⟹h=0+21gt2</p><p></p><p>\rm \implies\: h = \dfrac{1}{2} g {t}^{2}⟹h=21gt2</p><p></p><p>\rm \implies\:t = \sqrt{ \dfrac{2h}{g} }⟹t=g2h</p><p></p><p>Now, range will be :</p><p></p><p>\rm \: R = u_{x}(t)R=ux(t)</p><p></p><p>\rm \implies\: R = v \times t⟹R=v×t</p><p></p><p>\rm \implies\: R = v \times \sqrt{ \dfrac{2h}{g} }⟹R=v×g2h</p><p></p><p>\rm \implies\: 80 = v \times 4⟹80=v×4</p><p></p><p>\rm \implies\:v = 20 \: m {s}^{ - 1}⟹v=20ms−1</p><p></p><p>So, Velocity of projection is 20 m/s.</p><p></p><p>$