A ball is thrown vertically upward with a
speed of 25.4 m/s from a height of 2.1 m. (Remember gravity is a=9.8m/s^2)
How long does the ball take to hit the ground after it reaches its highest point?
Answer in units of s.
make a parobala, find the highest point find the time for the whole thing, then dive the time in half. hope that kind of helped !
height Y = 2.1 + 25.5 t - 4.9 t^2
Solve for the time Y = 0
It reaches its highest point when t = Vo/g = 25.4/9.8 = 2.59 s
Since they want the time it takes to hit the ground AFTER reaching the highest point, you will need to subtract 2.59s from the time Y=0
There is another way to do this without solvoing a quadratic equation>
1) use conservation of energy to calculate how much higher high it goes.
g*(deltaY) = Vo^2/2
2) Add the initial elevation, 2.1 m, to delta Y. That will give you Ymax
3) T (time to fall from max height)
= sqrt (2 Ymax/g)
1,\m/s
To find the time it takes for the ball to hit the ground after reaching its highest point, we need to break down the motion of the ball into two parts: the upward motion and the downward motion.
First, let's find the time it takes for the ball to reach its highest point during the upward motion. We can use the equation for vertical displacement:
s = ut + (1/2)at^2
Here, s is the vertical displacement (2.1 m), u is the initial velocity (25.4 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
Plugging in the values, we get:
2.1 = 25.4t + (1/2)(-9.8)t^2
Simplifying the equation, we have:
-4.9t^2 + 25.4t + 2.1 = 0
Now we can solve this quadratic equation to find the time t when the ball reaches its highest point. We can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -4.9, b = 25.4, and c = 2.1. Plugging in these values, we have:
t = (-25.4 ± √(25.4^2 - 4(-4.9)(2.1))) / (2(-4.9))
Simplifying further, we get:
t = (-25.4 ± √(645.16 + 40.92)) / (-9.8)
t = (-25.4 ± √686.08) / (-9.8)
Calculating the values inside the square root, we have:
t = (-25.4 ± 26.186) / (-9.8)
Now, solving for both values of t, we have:
t₁ = (-25.4 + 26.186) / (-9.8) ≈ 0.082 s
t₂ = (-25.4 - 26.186) / (-9.8) ≈ -5.361 s
Since time cannot be negative in this context, we discard the negative root and only consider t₁.
Therefore, the time it takes for the ball to reach its highest point is approximately 0.082 seconds.
To find the total time for the ball to hit the ground after reaching its highest point, we know that the upward and downward times are equal. So the total time is twice the time it took to reach the highest point.
Therefore, the total time for the ball to hit the ground after reaching its highest point is approximately 2 * 0.082 seconds, which is approximately 0.164 seconds.