Use a(t) = -32 ft/sec2 as the acceleration due to gravity. (Neglect air resistance.)
A ball is thrown vertically upward from a height of 4 feet with an initial velocity of 79 feet per second. How high will the ball go? (Round your answer to two decimal places.)
h = -16t^2 + 79t + 4
we need the vertex of this downwards parabola
the t of the vertex = -b/2a
= -79/-32
= 79/32 sec
for that time
height = -16(79/32)^2 + (79/32)(79) + 4
= 197.796..
= 197.80 correct to 2 decimals
check my arithmetic
To find the height the ball will reach, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 ft/s when ball reaches the highest point)
u = initial velocity (79 ft/s)
a = acceleration due to gravity (-32 ft/s^2)
s = distance traveled (height reached by the ball)
Substituting the given values into the equation:
0^2 = 79^2 + 2(-32)s
0 = 6241 - 64s
Rearranging the equation:
64s = 6241
Dividing by 64:
s = 6241/64
Calculating the value:
s ≈ 97.52
Therefore, the ball will reach a height of approximately 97.52 feet.
To find the maximum height reached by the ball, we can use the equation of motion for vertical motion under constant acceleration. The equation is given by:
h(t) = h0 + v0t + (1/2)at^2
where:
h(t) is the height of the ball at time t,
h0 is the initial height of the ball,
v0 is the initial velocity of the ball,
a is the acceleration due to gravity, and
t is the time.
Given:
h0 = 4 ft (initial height)
v0 = 79 ft/s (initial velocity)
a = -32 ft/s^2 (acceleration due to gravity)
We want to find the maximum height, so we need to determine the time it takes for the ball to reach that height. At the maximum height, the vertical velocity of the ball becomes zero. We can use this information to find the time it takes to reach the maximum height.
v(t) = v0 + at
At the maximum height, v(t) = 0, so we have:
0 = v0 + at
Now, let's solve this equation for t:
t = -v0 / a
Substituting the given values:
t = -79 ft/s / (-32 ft/s^2)
t ≈ 2.469 seconds
Now that we have the time it takes for the ball to reach the maximum height, we can substitute this value into the equation of motion to find the height.
h(t) = h0 + v0t + (1/2)at^2
h(2.469) = 4 ft + 79 ft/s * 2.469 s + (1/2)(-32 ft/s^2)(2.469 s)^2
h ≈ 4 ft + 194.551 ft + (-20.083 ft)
h ≈ 178.468 ft
Therefore, the ball will reach a maximum height of approximately 178.47 feet.