the height of a ball t seconds after it is thrown into the air from the top of the building can be modelled by h(t)= -16t2+48t+64 where h(t) is height in metres.how high is the building? how high does the ball rise before starting to drop downword, and after how many seconds does the ball hit the ground?
first of all, your equation gives the height in feet (on earth, anyway) because in metric units g = 9.81 m/s^2
But anyway, you know that the vertex of that parabola lies at t = 48/32 = 1.5
so find h(1.5)
it hits the ground when the height is zero, so solve
-16t^2+48t+64
or, after dividing by -16,
t^2 - 3t - 4 = 0
To find the height of the building, we need to determine the initial height (when t = 0) in the given equation. Let's calculate h(0):
h(t) = -16t^2 + 48t + 64
h(0) = -16(0)^2 + 48(0) + 64
= 64
Therefore, the height of the building is 64 meters.
The ball will reach its maximum height at the vertex of the parabolic function. The vertex can be found using the formula t = -b/2a. In this case, a = -16 and b = 48:
t = -b/2a
= -48 / (2*(-16))
= -48 / (-32)
= 1.5
So, the ball reaches its maximum height 1.5 seconds after being thrown into the air.
Next, we substitute t = 1.5 back into the original equation to find the height at that time:
h(t) = -16t^2 + 48t + 64
h(1.5) = -16(1.5)^2 + 48(1.5) + 64
= -16(2.25) + 72 + 64
= -36 + 72 + 64
= 100
Therefore, the ball rises to a height of 100 meters before starting to drop downward.
Lastly, to find the time it takes for the ball to hit the ground, we need to solve the equation h(t) = 0:
-16t^2 + 48t + 64 = 0
We can simplify the equation by dividing all terms by -16, leading to:
t^2 - 3t - 4 = 0
This equation can be factored as:
(t - 4)(t + 1) = 0
From this, we find that t = 4 seconds or t = -1 second. Since time cannot be negative in this context, the ball hits the ground after 4 seconds.
Therefore, the ball hits the ground after 4 seconds.
To find the height of the building, we need to determine the initial height from which the ball was thrown. In the given function, h(t) = -16t^2 + 48t + 64, t represents time in seconds and h(t) represents the height of the ball at that time.
We can observe that when the ball is initially thrown, the height at time t = 0 is h(0). Substituting t = 0 into the given equation, we can find the initial height of the ball:
h(0) = -16(0)^2 + 48(0) + 64
= 0 + 0 + 64
= 64
Therefore, the building is 64 meters high.
To determine the highest point the ball reaches before it starts descending, we need to find the vertex of the parabolic function h(t) = -16t^2 + 48t + 64.
The vertex of a parabola in the form y = ax^2 + bx + c is given by the equation x = -b/2a. In our case, a = -16 and b = 48.
t = -48 / 2(-16)
t = -48 / -32
t = 1.5
The ball reaches its highest point after 1.5 seconds.
Substituting t = 1.5 into the equation h(t) = -16t^2 + 48t + 64, we can find the height at that time:
h(1.5) = -16(1.5)^2 + 48(1.5) + 64
= -16(2.25) + 72 + 64
= -36 + 72 + 64
= 100
Therefore, the ball reaches a maximum height of 100 meters before it starts descending.
Now, let's find out when the ball hits the ground. To do this, we set h(t) = 0 and solve for t:
-16t^2 + 48t + 64 = 0
This is a quadratic equation that can be solved using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
t = (-48 ± √(48^2 - 4(-16)(64))) / (2(-16))
t = (-48 ± √(2304 + 4096)) / (-32)
t = (-48 ± √(6400)) / (-32)
t = (-48 ± 80) / (-32)
Therefore, the possible values of t are:
t1 = (-48 + 80) / (-32) ≈ 1.75
t2 = (-48 - 80) / (-32) ≈ -0.25
Since time cannot be negative in this context, we take t = 1.75 as the approximate time in seconds when the ball hits the ground.
To summarize:
- The height of the building is 64 meters.
- The ball rises to a height of 100 meters before starting to descend.
- The ball hits the ground after approximately 1.75 seconds.