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

510 m

Step-by-step explanation:

The arrow is fired upward, so the motion of the arrow is a free fall motion, therefore we can find the maximum height by using the suvat equation

`v^2-u^2=2as`

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have:

v = 0 (at the point of maximum height, the velocity of the arrow is zero)

u = 100 m/s (initial velocity)

`a=g=-9.8 m/s^2` (acceleration of gravity, downward)

s = ? is the maximum height

Solving for s, we find the maximum height of the arrow:

`s=(v^2-u^2)/(2a)=(0-100^2)/(2(9.8))=510 m`