Question

An arrow is launched upwards with an initial velocity of +48.0 m/s. Assuming that it was launched upwards from the ground, what will be the height of the arrow after 4.5 s?solve algebraically first then substitute in numbers and solve

Answer 1

Given:

The initial velocity of the arrow in an upward direction is

`v_o=48\text{ m/s}`

The time duration is t = 4.5 s.

Required: The distance travelled by the arrow.

Step-by-step explanation:

The distance travelled in an upward direction can be calculated by the formula,

`\begin{gathered} \Delta y=v_ot+(1)/(2)at^2 \\ \Delta y= v_(o)t-(1)/(2)gt^(2) \end{gathered}`

Here, a= -g is the acceleration due to gravity whose value is

`g=\text{ 9.8 m/s}^2`

On substituting the values, the distance covered by the arrow will be

`\begin{gathered} \Delta y=\text{ 48}*(4.5)-(1)/(2)*9.8*(4.5)^2 \\ =116.78\text{ m} \end{gathered}`

Final Answer: The distance covered by the arrow is 116.78 m