This about straight Motion

Question

Amanda wanted to throw an object into the window of her house. The window was 10 meters high

where Amanda was standing. Amanda's initial position stood 8 meters from the wall of the house. Amanda picked up

object up to 2 m from the ground and start throwing at a rate of 10 m / s. If it's Amanda

throwing an object at an angle of 53 ° to the horizontal, can the object be thrown inside

her window?

Answer 1

This question is about parabolic motion (e.g. the throwing of a ball in an arc in real life, subject to gravity) Start by finding the time it takes the ball to go horizontally from Amanda to the wall, 8 meters away.

change in x position at constant velocity (since gravity only acts in the y direction) would be:

Δx=v_xΔt

v_x=v₀cos(θ) where θ is the angle to the horizontal, and v₀ is the initial speed of the ball.

so v_x=10cos(53)

Δt=Δx/v_x=8/(10cos(53))≈1.329 seconds

Now we can solve for the resulting y coordinate after change in time Δt to find out where the ball will be after Δt seconds.

Based on the problem description, it looks like the angle and initial velocity are fixed, but the starting height y₀ can be anywhere between 0 and 2 meters. The desired final y is at 10 meters, so the resulting Δy must be between 10-2 and 10-0 meters, or:

8≤Δy≤10

Now to find Δy:

Δy=v₀sin(θ)*Δt-(g/2)(Δt)²

Δy=10sin(53)*Δt-(9.81/2)(Δt)²

Plug in Δt=8/(10cos(53)) and the result is:

Δy=1.95 meters

The highest final y at the house wall she could achieve with this throw is 3.95 meters (2m+Δy), not even close to the window.