If an object is propelled upward from ground level with initial velocity of 55 feet per second, it's height h in feet t seconds later is given by the equation H=-16t^2+55t. after how many seconds does the object hit ground?
V = Vo + g*Tr = 0 @ max Ht.
Tr = -Vo/g = -55/-32 = 1.72 s. = Rise time.
Tf = Tr = 1.72 s. = Fall time.
Tr+Tf = 1.72 + 1.72 = 3.44 s. To hit gnd
To find out after how many seconds the object hits the ground, we need to determine the time when the height (h) of the object equals zero.
Given that the equation for the height (h) is H = -16t^2 + 55t, we can set it equal to zero:
0 = -16t^2 + 55t
Now, let's solve this quadratic equation to find the times when the height is zero.
First, we can factor out t:
t(0 = -16t + 55)
Now we can solve for t by setting each factor equal to zero:
t = 0 (Equation 1)
0 = -16t + 55 (Equation 2)
Solving Equation 2 for t:
16t = 55
t = 55/16
Since negative time is not possible in this context, we disregard Equation 1 and consider t = 55/16.
Therefore, the object hits the ground after approximately 3.44 seconds (55/16 ≈ 3.44 seconds).