Hello,
A projectile is launched straight up from the ground with an initial velocity of 320 ft/sec.
a. When will the projectiles height above the ground be 1024 ft?
For this I got 16 seconds
b. When will the height of the projectile be at least 24 ft?
I am not sure how to figure this one out.
c. When will the object strike the ground?
I got 20 seconds
d. What is the maximum height of the projectile?
I got 1600 feet.
So, I am still unsure about b and I wanted to know if my other answers seem correct
Thanks
c) and d) are correct
for a) and the others I assume you used Calculus and the equation
distance = -16t^2 + 320t to find your answers
you must have set 1024 = -16t^2 + 320t
after re-arranging and dividing by 16 this would give you
t^2 - 20t + 64 = 0
(t-16)(t-4) = 0
so t = 16 or t = 4
you had the t=16, but failed to show the t=4 case.
It reached a height of 1024 twice, at 4 seconds after liftoff on its way up, and then 16 seconds later on its way down
for b)
set 24 = -16t^2 + 320t
16t^2 - 320t + 24 = 0
2t^2 - 40t + 3 = 0
using the quadratic formula
t = (40 ± √(1600-4(2)(3))/4
= 19.92 or .075
so between the times of .075 and 19.92 seconds the object is at least 24 feet high.
My mistake. b should have asked at least 1024 ft. So does that mean the answer would be anytime in between 4 and 16 seconds?
b. When will the height of the projectile be at least 1024 ft?
Let's go through each part of the question step by step:
a. To find when the projectile's height above the ground is 1024 ft, we can use the equation for vertical displacement of a projectile:
y = v0*t - (1/2)*g*t^2
where:
y is the height above the ground,
v0 is the initial velocity,
g is the acceleration due to gravity (32 ft/s^2),
and t is the time.
Plugging in the given values, we have:
1024 = 320*t - (1/2)*32*t^2
Simplifying, we get a quadratic equation:
16t^2 - 320t + 1024 = 0
Factoring the equation, we have:
16(t - 8)(t - 8) = 0
Solving for t, we find that t = 8 seconds.
Therefore, the projectile's height above the ground will be 1024 ft at 8 seconds.
b. To find when the height of the projectile is at least 24 ft, we need to determine the time it takes for the projectile to reach its highest point. At this point, the velocity will be 0. Using the equation for vertical velocity, we have:
v = v0 - g*t
Setting v = 0, we can solve for t:
0 = 320 - 32*t
Solving for t, we find that t = 10 seconds.
Since the projectile starts from the ground, it will pass the 24 ft mark on its way up, so the height of 24 ft will be reached before 10 seconds.
c. The time for the object to strike the ground can be found by setting the height, y, equal to 0. Using the equation for vertical displacement, we have:
0 = 320t - (1/2)*32*t^2
This simplifies to:
16t^2 - 320t = 0
Factoring out t, we find:
16t(t - 20) = 0
So t = 0 or t = 20 seconds.
Since the object was launched from the ground, we can discard t = 0. Thus, the object will strike the ground at 20 seconds.
d. The maximum height of the projectile can be found by determining the time it takes for the vertical velocity to reach zero. At this point, the projectile is at its highest point. Using the equation for vertical velocity, we have:
v = v0 - g*t
Setting v = 0, we can solve for t:
0 = 320 - 32*t
Solving for t, we find that t = 10 seconds.
Plugging this value of t into the equation for vertical displacement, we have:
y = 320*10 - (1/2)*32*(10)^2
Simplifying, we get:
y = 1600 ft.
Therefore, the maximum height of the projectile is 1600 ft.
To summarize:
a. The projectile's height above the ground will be 1024 ft at 8 seconds.
b. The height of the projectile will be at least 24 ft before it reaches its highest point at 10 seconds.
c. The object will strike the ground at 20 seconds.
d. The maximum height of the projectile is 1600 ft.
To find the answers to these questions, we can use basic kinematic equations. Let's go through each question one by one:
a. When will the projectile's height above the ground be 1024 ft?
To find the time it takes for the projectile to reach a height of 1024 ft, we can use the equation:
s = u*t - (1/2)*g*t^2,
where s is the height, u is the initial velocity, g is the acceleration due to gravity (32 ft/sec^2), and t is the time.
Rearranging the equation to solve for t:
1024 = 320*t - (1/2)*32*t^2
0 = 16t^2 - 320t + 1024
Now we can solve this quadratic equation. We can factor it or use the quadratic formula. Factoring gives us:
0 = 16(t-16)(t - 4)
So, t = 16 or t = 4.
Since we are looking for the time it takes to reach a height of 1024 ft, we know that it must be greater than the initial launch time. Therefore, t = 16 seconds is the correct answer.
b. When will the height of the projectile be at least 24 ft?
To find the time when the height is at least 24 ft, we need to solve the equation similarly:
24 = 320*t - (1/2)*32*t^2
0 = 16t^2 - 320t + 768
Again, we can factor or use the quadratic formula. Factoring gives us:
0 = 16(t-24)(t-2)
So, t = 24 or t = 2.
Since we are looking for the time when the height is at least 24 ft, we take the smaller value, which is t = 2 seconds.
c. When will the object strike the ground?
To find the time when the object strikes the ground, we need to consider that at the highest point of the projectile's path, the height will be zero. So, using the equation:
0 = 320*t - (1/2)*32*t^2
We can arrange this as a quadratic equation:
0 = 16t^2 - 320t
Factoring:
0 = 16t(t - 20)
So, t = 20 or t = 0.
Since time cannot be negative, t = 20 seconds is the correct answer.
d. What is the maximum height of the projectile?
The maximum height is reached when the vertical velocity becomes zero. To find this height, we can determine the time it takes to reach the maximum height. The formula for vertical velocity is:
v = u - g*t,
where v is the vertical velocity and u is the initial velocity.
Setting v = 0, we have:
0 = 320 - 32*t
So, t = 10 seconds.
Now, we can use this value of t in the equation for height:
s = u*t - (1/2)*g*t^2
s = 320*10 - (1/2)*32*(10^2)
s = 3200 - 1600
s = 1600 ft.
So, the maximum height of the projectile is 1600 ft.
In summary, the correct answers are:
a. 16 seconds
b. 2 seconds
c. 20 seconds
d. 1600 feet