the distance travelled by an object thrown upwards in t seconds is 30t - 4.9t square metres. after how much time would it fall down? at what all times would it be 20 metres above the ground?
When it falls down, the distance is zero
so 0 = 30t - 4.9t^2
t(30 - 4.9t) = 0
t = 0 , that would be at the beginning
or
t = 30/4.9 or appr. 6.12 seconds
2nd part:
20 = 30t - 4.9t^2
4.9t^2 - 30t + 20 = 0
t = (30 ± √508)/9.8
= .761 seconds, on its way up
or
= 5.36 seconds, on the way down
To determine when the object would fall down, we need to find the time at which the height becomes zero.
Given that the height (h) of the object is given by the equation h = 30t - 4.9t^2, we can set this equation equal to zero and solve for t:
30t - 4.9t^2 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, factoring is simpler. We can rewrite the equation as:
t(30 - 4.9t) = 0
Setting each factor equal to zero, we have:
t = 0 (this represents the initial time when the object is thrown)
30 - 4.9t = 0
Solving the second equation for t:
4.9t = 30
t = 30/4.9 ≈ 6.12 seconds
Therefore, the object would fall down after approximately 6.12 seconds.
Now, let's determine the times at which the object is 20 meters above the ground. We need to find the values of t for which the height is 20 meters.
Setting the height equation equal to 20:
30t - 4.9t^2 = 20
To solve this quadratic equation, we subtract 20 from both sides:
4.9t^2 - 30t + 20 = 0
This equation can also be factored or solved using the quadratic formula. Let's use factoring this time:
(4.9t - 10)(t - 2) = 0
Setting each factor equal to zero:
4.9t - 10 = 0 or t - 2 = 0
Solving each equation separately:
4.9t = 10
t = 10/4.9 ≈ 2.04 seconds
t = 2
Therefore, the object would be 20 meters above the ground at approximately 2.04 seconds and also at 2 seconds.