Question

A father demonstrates projectile motion to his children by placing a pea on his fork's handle and rapidly depressing the curved tines, launching the pea to heights above the dining room table. Suppose the pea is launched at 7.47 m/s at an angle of 81.0° above the table. With what speed (in m/s) does the pea strike the ceiling 2.10 m above the table?

Answer 1

Given

Initial velocity is u=7.47 m/s

The angle is

`\theta=81^0`

The vertical height travelled, h=2.10 m

To find

The final speed of the pea.

Step-by-step explanation

Along the vertical direction,

The final speed is given by the equation of kinematic

`v_y^2=u_y^2+2ah`

Putting the values

`\begin{gathered} v_y^2=(7.47sin81^o)^2+2(-9.81)*2.1 \\ \Rightarrow v_y=3.63\text{ m/s} \end{gathered}`

Along the horizontal direction,

the speed is constant throughout the projectile

Thus,

`v_x=7.47cos81^0=1.168\text{ m/s}`

So the final velocity of the pea is

`\begin{gathered} v=√((3.63)^2+(1.168)^2) \\ \Rightarrow v=3.81\text{ m/s} \end{gathered}`

Conclusion

The final velocity is 3.81 m/s