Question

An arrow is launched upward with an initial speed of 100 meters per second (m/s). The equations above describe the constant-acceleration motion of the arrow, where v0 is the initial speed of the arrow, v is the speed of the arrow as it is moving up in the air, h is the height of the arrow above the ground, t is the time elapsed since the arrow was projected upward, and g is the acceleration due to gravity (9.8 m/s2 ). What is the maximum height from the ground the arrow will rise to the nearest meter?

Answer 1

Answer:

Maximum height, h = 510 meters

Step-by-step explanation:

Initial speed of the arrow, $v_o=100\ m/s$

The equations that describes the constant-acceleration motion of the arrow are as follows :

$v=v_o-gt$ (speed -time)

$h=v_ot-\dfrac{1}{2}gt^2$ (position- time)

$v^2=v_o^2-2gh$ (position- speed)

Where

v₀ is the initial speed of the arrow

v is the speed of the arrow as it is moving up in the air

h is height of the arrow above the ground

t is the time elapsed since the arrow was projected upward

g is the acceleration due to gravity

We can use the third equation to find the maximum height from the ground the arrow will rise. At maximum height, v = 0

So, $0=v_o^2-2gh$

$h=\dfrac{v_o^2}{2g}$

$h=\dfrac{(100)^2}{2\times 9.8}$

h = 510.20 meters

or

h = 510 meters

So, the maximum height from the ground the arrow will rise to 510 meters. Hence, this is the required solution.