When a football is kicked with a vertical speed of 20m/s its height h, in meters, after t seconds is given by the equation h = -5t&2 +20t
a ) how long is the football in the air
b ) how long is it kicked is the football at a height of 15 m
How am i supposed to solve this
I meant h= - 5^2+20t
a) h = 0 when the ball lands ... 0 = -5 t^2 + 20 t ... solve for t
b) is unintelligible
To solve these questions, you can use the given equation for the height of the football in terms of time:
h = -5t^2 + 20t
a) To find how long the football is in the air, you need to determine when the height is equal to zero. At the highest point of the ball's trajectory, the height is zero. So, set the equation equal to zero:
0 = -5t^2 + 20t
Now, you can solve this quadratic equation. Factor out a common term of "t":
0 = t(-5t + 20)
This equation is equal to zero when either t = 0 or -5t + 20 = 0.
If t = 0, it means the ball was at the ground initially, which is not what we are looking for. Therefore, solve -5t + 20 = 0:
5t = 20
t = 20/5
t = 4
So, the football is in the air for 4 seconds.
b) To find how long the football is kicked when it is at a height of 15 meters, substitute the height h with 15 in the equation:
15 = -5t^2 + 20t
Rearrange this equation to form a quadratic equation:
5t^2 - 20t + 15 = 0
Now solve this quadratic equation to find the values of t. Factoring may be the easiest method for this equation:
5t^2 - 15t - 5t + 15 = 0
5t(t - 3) - 5(t - 3) = 0
(5t - 5)(t - 3) = 0
Setting each factor equal to zero:
5t - 5 = 0
5t = 5
t = 5/5
t = 1
t - 3 = 0
t = 3
So, the football is kicked at the height of 15 meters for two moments: at 1 second and 3 seconds.
To solve the given equations, we will need to analyze the height-time relationship provided.
a) To find out how long the football is in the air, we need to determine when the height (h) of the football is zero. We can do this by setting h = 0 and solving for t in the given equation:
h = -5t² + 20t
0 = -5t² + 20t
To simplify the equation, let's divide the entire equation by -5:
0 = t² - 4t
Now, let's factorize the equation:
0 = t(t - 4)
From here, we have two possibilities:
1) t = 0: This means that when the football is initially kicked, time (t) is equal to zero.
2) t - 4 = 0: This means the football is on the ground at t = 4 seconds.
However, we are only interested in the time that the football is in the air, so we disregard t = 0. Therefore, the football is in the air for a duration of 4 seconds.
b) To find out how long the football is at a height of 15 m, we need to set h = 15 and solve for t in the given equation:
h = -5t² + 20t
15 = -5t² + 20t
Rearranging the equation, we get:
5t² - 20t + 15 = 0
Dividing the equation by 5:
t² - 4t + 3 = 0
Now, let's factorize the equation:
(t - 1)(t - 3) = 0
This gives us two possibilities:
1) t - 1 = 0: This means the football reaches a height of 15 m at t = 1 second.
2) t - 3 = 0: This means the football reaches a height of 15 m at t = 3 seconds.
Therefore, the football is kicked for a duration of either 1 or 3 seconds to reach a height of 15 meters, depending on the specific scenario.