Question

2. A javelin is thrown through an angle of 30o to the horizontal. Find the time taken by the javelin to reach the maximum height if the initial velocity is 60ms-1 [ take g =10 ms-1]

Answer 1

3 seconds

Step-by-step explanation:

The initial velocity is 60 m/s at an angle of 30° to the horizontal, so we can calculate the initial velocity in the vertical direction as

Viy = Vi sin(θ)

Viy = 60 sin(30)

Viy = 60(0.5)

Viy = 30 m/s

Then, we can use the following equation to calculate the time that it takes to reach the maximum height.

`\begin{gathered} v_(fy)=v_(iy)-gt \\ v_(fy)+gt=v_(iy) \\ gt=v_(iy)-v_(fy) \\ \\ t=(v_(iy)-v_(fy))/(g) \end{gathered}`

At a maximum height the vertical velocity is 0 m/s, so replacing viy = 30 m/s, vfy = 0 m/s and g = 10 m/s², we get

`\begin{gathered} t=\frac{30\text{ m/s-0m/s}}{10\text{ m/s}^2} \\ \\ t=\frac{30\text{ m/s}}{10\text{ m/s}^2} \\ \\ t=3\text{ s} \end{gathered}`

Therefore, the answer is 3 seconds