Question

A lad wants to throw a bag into the open window of his friend's room 10.0 m above. Assuming itjust reaches the window, he throws the bag at 60.0° to the ground:a) At what velocity should he throw the bag? (16.2 m/s at 60.0° to the ground]b) How far from the house is he standing when he throws the bag? (11.5 m]

Answer 1

Given data:

* The height of the window is 10 m.

* The angle of the incident velocity with respect to the horizontal is 60 degree.

Solution:

(a). The height of the projectile in terms of the initial velocity is,

`H=(u^2\sin ^2(\theta))/(2g)`

where u is the initial velocity, g is the acceleration due to gravity, and

`\theta\text{ is the angle made by initial velocity with the horizontal}`

Substituting the known values,

The initial velocity of the projectile is,

`\begin{gathered} 10=(u^2\sin ^2(60^(\circ)))/(2*9.8) \\ u^2=(10*2*9.8)/(\sin ^2(60^(\circ))) \\ u^2=261.33 \\ u=16.16ms^(-1) \\ u\approx16.2ms^(-1) \end{gathered}`

Thus, the initial velocity of the bag is 16.2 m/s at an angle of 60 degree to the ground.

(b). The horizontal range of bag is,

`H=(u^2\sin (2\theta))/(g)`

Substituting the known values,

`\begin{gathered} H=(16.16^2*\sin(2*60))/(9.8) \\ H=23.077\text{ m} \end{gathered}`

The position of the bag from the house is,

`\begin{gathered} R^(\prime)=(H)/(2) \\ R^(\prime)=11.53\text{ m} \\ R^(\prime)\approx11.5\text{ m} \end{gathered}`

Thus, the lad is standing at the distance of 11.5 m from the house.