Question

An arrow is launched straight up from the ground with an initial velocity of 23.4 m/s. How long until it reaches its highest point?

Answer 1

Given:-

• Initial Velocity = 23.4m/s

• Final Velocity = 0m/s ( Highest Point )

• Acceleration due to gravity = -9.8m/s²

To Find:-

• Height it reaches.

Formulae used:-

• v² - u² = 2gh

• h = ut + ½ × g × t²

Where,

• h = Distance
• u = Initial Velocity
• g = Acceleration
• t = Time
• v = Final Velocity

Now,

→ v² - u² = 2gh

→ (0)² - (23.4)² = 2 × -9.8 × h

→ -547.56 = -19.6h

→ h = -547.56/-19.6

→ h = 27.93m

Therefore,

→ h = ut + ½ × g × t²

→ 27.93 = 23.4t + ½ × -9.8 × t²

→ 27.93 = 23.4t - 4.9t²

→ t = 2.35s

Hence, The Arrow will take 2.35s to reach the highest Point.

Answer 2

Given : -

• The initial velocity of an arrow is 23.4 m/s

To Find : -

• How long until it reaches its highest point?

Solution : -

Using third equation of motion to find height first.

$\sf : \implies \: v^2-u^2=2ah$

At maximum height, v = 0

a = - g

$\sf: \implies \: u^2=2gh\\\\ \sf : \implies \: h=\dfrac{u^2}{2g}\\\\ \sf : \: \implies \: h=\dfrac{(23.4)^2}{2\times 9.8}\\\\ \sf : \: \implies \: h=27.93\ m$

Let t is the time to reach its maximum height. So, it can be calculated using second equation motion as follows :

$\sf : \implies \: 27.93=23.4t-\dfrac{1}{2}\cdot9.8\cdot t^{2}\\\\ \: \sf : \implies \: t=2.35\ s$

Hence, it will take 2.35 seconds to reach its maximum height.