A ball is thrown from the top of a 50-ft building
with an upward velocity of 24 ft/s. When will it
reach its maximum height? How far above the
ground will it be?
Using your data,
h = -16t^2 + 24t + 50
This is a downwards opening parabola, so find its vertex and all the
mysteries will be revealed.
the model can be written this way:
height= vi*t + hi -16t^2 where hi=50ft; vi=24ft/sec
so this is a parabola. The zeroes can be found
0=23*t+50-16t^2
16t^2-23t-50=0
t=(23+-sqrt(23^2+4*16*50))/32
solving that, t=2.63, -1.15 so the max height occurs halfway between these zeroes
tmax=3.78/2-1.15= .74 sec
hmax=h(.74)=height= vi*t + hi -16t^2
To find out when the ball will reach its maximum height, we need to use the equation of motion for vertical motion. The equation is:
h = ht + (v₀t) - (1/2)gt²
Where:
h = height from the ground
ht = initial height (50 ft in this case)
v₀ = initial velocity (24 ft/s in this case)
g = acceleration due to gravity (32.2 ft/s²)
At the maximum height, the vertical velocity becomes zero (v = 0). So, we can set the equation equal to zero and solve for the time (t):
0 = ht + (v₀t) - (1/2)gt²
0 = 50 + 24t - (1/2)(32.2)t²
Now we can solve this quadratic equation for t:
0 = -16.1t² + 24t + 50
To find the time when the ball reaches its maximum height, we need to find the value of t that gives a height of zero. Using the quadratic formula, we have:
t = (-b ± √(b² - 4ac)) / (2a)
Here, a = -16.1, b = 24, and c = 50. Plugging in these values, we can solve for t.