An object is thrown upward with a velocity of 128ft/sec.Find out the time taken to reach the ground.
To find the time taken for the object to reach the ground, we can use the equation of motion for free-falling objects. The equation is:
y = y0 + v0 * t - (1/2) * g * t^2
Where:
y = final position (in this case, the ground)
y0 = initial position (height from which the object is thrown, which is usually taken as 0 in these problems)
v0 = initial velocity (velocity with which the object is thrown)
t = time taken
g = acceleration due to gravity (approximately 32 ft/sec^2)
Since the object is thrown upward, the initial velocity will be positive. We can substitute the given values into the equation and solve for time (t):
y = 0 + 128 * t - (1/2) * 32 * t^2
0 = 128t - 16t^2
Now, we have a quadratic equation. To solve for t, we can factor or use the quadratic formula.
In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
For our equation: 0 = -16t^2 + 128t
a = -16
b = 128
c = 0
Substituting in the values:
t = (-128 ± √(128^2 - 4 * -16 * 0)) / (2 * -16)
t = (-128 ± √(16384)) / -32
t = (-128 ± 128) / -32
Now, solving for t, we have two possible solutions:
t1 = (-128 + 128) / -32 = 0
t2 = (-128 - 128) / -32 = 8
The first solution (t = 0) represents the initial time when the object is thrown, so we ignore it. Thus, the time taken for the object to reach the ground is 8 seconds.
you know that the height
h = 128t - 16t^2 = 16t(8-t)
so, note that it hits the ground when h=0.
now just do the math.