You shoot an arrow into the air. 2.oo seconds later the arrow has gone straight upward to a height of 30 meters above its launch point. (a) What was the arrow's initial speed? (b) How long did it take for the arrow to first reach a height of 15 meters above it's launch point?
Y = Vo*t -(g/2)t^2
At t = 2.00s,
30 = 2*Vo -(4.9)*4
(a) Vo = 24.8 m/s
(b) Now that you know Vo, use the first equation again to solve for t when Y = 15 m.
3.45
5.6
To find the answers to these questions, we can use the equations of motion for an object moving vertically under constant acceleration due to gravity. These equations are:
1. Position equation: y = yo + vot - 0.5gt^2
2. Velocity equation: v = vo - gt
3. Final velocity equation: v^2 = vo^2 - 2g(y - yo)
Where:
- y is the vertical position of the object
- yo is the initial vertical position of the object
- v is the velocity of the object
- vo is the initial velocity of the object
- g is the acceleration due to gravity (approximately -9.8 m/s^2)
- t is the time
Let's start by solving part (a):
(a) What was the arrow's initial speed?
Given:
- The arrow reached a height of 30 meters above its launch point.
- The time elapsed is 2.00 seconds.
Using the position equation, we can set up the following equation:
30 = 0 + vo(2.00) - 0.5(9.8)(2.00)^2
Simplifying this equation will give us the value of the initial velocity vo.
Now let's move on to solving part (b):
(b) How long did it take for the arrow to first reach a height of 15 meters above its launch point?
Given:
- The arrow reached a height of 15 meters above its launch point.
Using the position equation, we can set up the following equation:
15 = 0 + vo(t) - 0.5(9.8)(t)^2
Again, solving this equation will give us the value of the time t.
By using these two equations and substituting the given values, we can find both the initial speed of the arrow and the time taken for it to reach a certain height.