Question

A penguin running ( maybe waddling) 1.6 m/s dives out horizontally from the edge of a glacier and reaches the water below 3.0s later. How high vertical face of the glacier and how far from its base did the penguin hit the water ?

Answer 1

Given data

*The given speed of the penguin is v = 1.6 m/s

*The given time is t = 3.0 s

*The value of the acceleration due to gravity is g = 9.8 m/s^2

The formula for the height of the vertical face of the glacier is given as

`h=(1)/(2)gt^2`

Substitute the known values in the above expression as

`\begin{gathered} h=(1)/(2)(9.8)(3.0)^2 \\ =44.1\text{ m} \end{gathered}`

Hence, the vertical face of the glacier is h = 44.1 m

The formula for the distance of the penguin from the base to hit the water is given as

`d=vt`

Substitute the known values in the above expression as

`\begin{gathered} d=(1.6)(3.0) \\ =4.8\text{ m} \end{gathered}`

Hence, the distance of the penguin from the base to hit the water is d = 4.8 m