A projectile is launched from a platform 10 meters above ground with an initial upward velocity of 20 meters per second.
Find the time when the projectile has returned to the initial height launch, find the time the projectile hits the ground, find the time when the projectile hits the ground when its velocity is 0, find the height of the projectile above ground after 1 second
g = -9.8 m/s^2
v = +20 - 9.8 t
y = 10 + 20 t -4.9 t^2
a) when is y = 10 again?
10 = 10 + 20 t -4.9 t^2
t(4.9 t-20) = 0
t = 20/4.9 = 4.08 seconds
b) when is y = 0 ?
4.9 t^2 - 20 t - 10 = 0
t = [20 +/- sqrt (400+196)] / 9.8
forget negative time, that was before we started, use + sign
t = 4.53 seconds
c) velocity is 0 at the op when
0 = 20 -9.8 t
t = 2.04 seconds
d) y = 10 + 20 (1) -4.9 (1)^2
= 25.1 meters
To solve these problems, we can use the laws of motion and equations of motion under constant acceleration.
1. Find the time when the projectile has returned to the initial launch height:
When the projectile is at the same height as the launch point, its displacement is zero. We can use the equation of motion for displacement:
s = ut + (1/2)at^2
where:
s = displacement (0 in this case),
u = initial velocity (20 m/s),
a = acceleration (-9.8 m/s^2, assuming the projectile experiences acceleration due to gravity),
t = time.
Rearranging the equation, we get:
0 = 20t + (1/2)(-9.8)t^2
Simplifying further:
0 = 20t - 4.9t^2
To solve this quadratic equation, we can factor it or use the quadratic formula. Factoring it, we have:
0 = t(20 - 4.9t)
This equation has two solutions: t = 0 and (20 - 4.9t) = 0.
Ignoring the t = 0 solution (as that is the initial launch time), we solve the equation (20 - 4.9t) = 0 for t:
t = 20/4.9
t ≈ 4.08 seconds
Therefore, the projectile takes approximately 4.08 seconds to return to its initial launch height.
2. Find the time the projectile hits the ground:
To find the time when the projectile hits the ground, we need to find the time it takes for the projectile to travel a vertical distance of 10 meters (equal to the height of the platform).
We can use the same equation for displacement, but now our initial displacement (s) is -10 meters (since the projectile is moving downwards):
-10 = 20t + (1/2)(-9.8)t^2
Simplifying this quadratic equation:
-9.8t^2 + 20t - 10 = 0
Solving this equation gives us two possible values of t. Ignoring the t = 0 solution again (as it represents the initial launch time), we find:
t ≈ 1.20 seconds and t ≈ 1.63 seconds
Therefore, the projectile takes approximately 1.20 seconds and 1.63 seconds to hit the ground from its launch point.
3. Find the time when the projectile hits the ground when its velocity is 0:
To find the time when the projectile hits the ground with a velocity of 0, we set the upward velocity to 0 (since the projectile is moving downwards). We can use the equation:
v = u + at
where:
v = final velocity (0 m/s),
u = initial velocity (20 m/s),
a = acceleration (-9.8 m/s^2),
t = time.
Rearranging the equation, we have:
0 = 20 - 9.8t
Solving for t:
9.8t = 20
t ≈ 2.04 seconds
Therefore, the projectile takes approximately 2.04 seconds to hit the ground when its velocity is 0.
4. Find the height of the projectile above the ground after 1 second:
To find the height of the projectile after 1 second, we can use the equation of motion for displacement once again. This time, our initial velocity is 20 m/s, and we substitute the value of t as 1 second:
s = ut + (1/2)at^2
Substituting the values, we have:
s = (20)(1) + (1/2)(-9.8)(1)^2
Simplifying this equation:
s = 20 - 4.9
s ≈ 15.1 meters
Therefore, after 1 second, the projectile is approximately 15.1 meters above the ground.