If a projectile is shot vertically upward from the ground with an initial velocity of 100 ft per sec, neglecting air resistance, its height s (in feet) above the ground t seconds after projection is given by s= -16t^2 + 100t
a) After how many seconds will it be 50ft above the ground ?
B) How long will it take for the projectile to return to ground ?
A) Plug in 50 for s in the equation.
50=-16t^2+100t
Move the 50 over.
0=-16t^2+100t+50
Use the quadratic formula to solve for t and neglect the negative answer. If you don't know the quadratic formula, it's:
(-b plus or minus the square root of b^2-4ac)/2a
B)Plug 0 in for s (since the ground represents 0 feet) and follow the same steps from A).
Thank you,,it is very helpful
a) To find the time it takes for the projectile to be 50ft above the ground, we need to set s equal to 50 and solve for t.
s = -16t^2 + 100t
50 = -16t^2 + 100t
Rearranging the equation, we have a quadratic equation:
-16t^2 + 100t - 50 = 0
To solve this equation, we can either factor or use the quadratic formula. Let's use the quadratic formula.
The quadratic formula is: t = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = -16, b = 100, and c = -50.
t = (-100 ± √(100^2 - 4(-16)(-50))) / (2(-16))
Simplifying, we have:
t = (-100 ± √(10000 - 3200)) / (-32)
t = (-100 ± √(6800)) / (-32)
t = (-100 ± 82.462) / (-32)
We get two possible solutions:
t = (-100 + 82.462) / (-32) = -0.58 seconds (ignoring the negative value since time cannot be negative)
t = (-100 - 82.462) / (-32) = 6.42 seconds
Therefore, it will be 50ft above the ground after approximately 6.42 seconds.
b) To find the time it takes for the projectile to return to the ground, we need to set s equal to 0 and solve for t.
s = -16t^2 + 100t
0 = -16t^2 + 100t
Rearranging the equation, we have:
-16t^2 + 100t = 0
Factoring out a common factor of 4t, we have:
4t(-4t + 25) = 0
Setting each factor equal to zero, we have:
4t = 0 or -4t + 25 = 0
Solving the first equation, we get:
t = 0
However, this is the initial time when the projectile was shot, so it does not count.
Solving the second equation, we have:
-4t + 25 = 0
-4t = -25
t = 25/4 = 6.25 seconds
Therefore, it will take approximately 6.25 seconds for the projectile to return to the ground.
To find the time it takes for the projectile to be at a certain height, we can set the equation for height equal to the desired height and solve for time. Let's solve each part of the question step by step.
a) To find the time it takes for the projectile to be 50 feet above the ground, we can set s (the height) equal to 50 and solve for t.
s = -16t^2 + 100t
Setting s = 50:
50 = -16t^2 + 100t
Now let's solve this quadratic equation:
-16t^2 + 100t - 50 = 0
Divide through by -2 to simplify:
8t^2 - 50t + 25 = 0
Unfortunately, this quadratic equation doesn't easily factor, so we'll use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values from our equation, a = 8, b = -50, c = 25:
t = (-(-50) ± √((-50)^2 - 4(8)(25))) / (2(8))
Simplifying further:
t = (50 ± √(2500 - 800)) / 16
t = (50 ± √(1700)) / 16
Calculating the square root of 1700:
t ≈ (50 ± 41.23) / 16
Now we have two solutions:
t ≈ (50 + 41.23) / 16 ≈ 5.2 seconds
t ≈ (50 - 41.23) / 16 ≈ 0.5 seconds
Therefore, the projectile will be 50 feet above the ground after approximately 0.5 seconds and 5.2 seconds.
b) To find how long it takes for the projectile to return to the ground, we need to find when the height (s) will be zero.
Setting s = 0:
0 = -16t^2 + 100t
Again, this is a quadratic equation. Let's solve it:
-16t^2 + 100t = 0
Divide through by 4t to simplify:
-4t + 25 = 0
Solving for t:
4t = 25
t = 25 / 4
t = 6.25 seconds
Therefore, the projectile will take approximately 6.25 seconds to return to the ground.