an archer at the edge of a cliff fires an arrow from a height of 14.6 m above the ground at the bottom of the cliff. the arrow just clears a rock on a pillar that is 21.7 m above the gound which is 45.2 m horizontally away from the archer.determine
a,the initial speed of the arrow
b,the angle above the horizontal that the arrow is fired at
c,how far past the rock on the pillar that the arrow lands
An arrow was fired from a 45.2m tall cliff and reached a distance of 802m.
To solve this problem, we can use projectile motion equations. Let's break down the problem into three parts: vertical motion, horizontal motion, and the time of flight.
a) Initial speed of the arrow (also known as the magnitude of the initial velocity):
- Since we have the vertical component of the initial velocity, we can find the magnitude using the formula: v₀ = √(v₀x² + v₀y²)
- v₀y = √(2 * g * h), where g is the acceleration due to gravity (approximately 9.8 m/s²) and h is the height (14.6 m)
- v₀y = √(2 * 9.8 * 14.6)
- Calculate v₀y.
b) Angle above the horizontal that the arrow is fired at:
- We can use trigonometry to find the angle. The tangent of the angle is given by: tan(θ) = v₀y / v₀x
- Rearranging the equation gives: θ = atan(v₀y / v₀x)
- Calculate θ using the inverse tangent function (atan).
c) Distance past the rock on the pillar that the arrow lands:
- We can calculate the time of flight using the formula: t = 2 * v₀y / g
- Once we know the time, we can find the horizontal distance using the formula: d = v₀x * t
- Calculate d by substituting the values of v₀x and t. The value you get will be the distance past the rock on the pillar that the arrow lands.
Now you have a step-by-step explanation of how to solve this problem. Plug in the given values and perform the calculations to find the answers.