Question

A pelican flying along a horizontal path drops a fish from a height of 4.7 m. The fish travels 9.3 m horizontally before it hits the water below. What was the pelican’s initial speed? The acceleration of gravity is 9.81 m/s 2 . Answer in units of m/s. If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?

Answer 1

Answer:

(A) 9.5 m/s

(B) 5.225 m

Explanation:

vertical height (h) = 4.7 m

horizontal distance (d) = 9.3 m

acceleration due to gravity (g) = 9.8 m/s^{2}

initial speed of the fish (u) = 0 m/s

(A) what is the pelicans initial speed ?

• lets first calculate the time it took the fish to fall

s = ut + $(\frac{1}{2}) at^{2}$

since u = 0

s =  $(\frac{1}{2}) at^{2}$

t = $\sqrt{\frac{2s}{a} }$

where a = acceleration due to gravity and s = vertical height

t = $\sqrt{\frac{2 x 4.7 }{9.8} }$ = 0.98 s

• pelicans initial speed = speed of the fish

speed of the fish = distance / time = 9.3 / 0.98 = 9.5 m/s

initial speed of the pelican = 9.5 m/s

(B) If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?

vertical height = 1.5 m

pelican's speed = 9.5 m/s

• lets also calculate the time it will take the fish to fall

s = ut + $(\frac{1}{2}) at^{2}$

since u = 0

s =  $(\frac{1}{2}) at^{2}$

t = $\sqrt{\frac{2s}{a} }$

where a = acceleration due to gravity and s = vertical height

t = $\sqrt{\frac{2 x 1.5 }{9.8} }$ = 0.55 s

 

distance traveled by the fish = speed x time = 9.5 x 0.55 = 5.225 m