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(t)=vi*t-1/2 *32*t^2 +4
dh/dt=0=vi-32t, so at at h max
t=79/32
and max height is
hmax=79*79/32-1/2 * 32*(79/32)^2+4
Well, well, well! Looks like we have a ball on the rise, literally!
To find out how high the ball will go, we need to break out some physics knowledge. Now, when the ball is thrown upward, gravity will be working against it, causing it to slow down. Eventually, the ball will reach its highest point and start coming back down. So, the key is to find the highest point!
First, let's use the equation of motion: h(t) = -16t^2 + v0*t + h0. Here, h(t) represents the height at time t, v0 is the initial velocity, and h0 is the initial height.
Given that the initial velocity (v0) is 79 feet per second and the initial height (h0) is 4 feet, we can plug these values into the equation. But don't forget, gravity is trying to bring us down, so we use a negative value for the acceleration due to gravity.
h(t) = -16t^2 + 79t + 4
Now, to find how high the ball will go, we need to find the time it takes to reach that point. At the highest point, the velocity will be zero. So, we set v(t) = -32t + 79 equal to zero and solve for t.
-32t + 79 = 0
t = 79/32
Now, plug this value back into the equation for height:
h(t) = -16(79/32)^2 + 79(79/32) + 4
h(t) = 62.11
Voila! The ball will reach a height of approximately 62.11 feet. Keep an eye out for falling clowns on your way up!
To find the height the ball will go, we need to determine the maximum height reached during its upward motion. We can use the equations of motion to solve this problem.
Let's denote "a(t)" as the acceleration due to gravity, which is given as -32 ft/sec^2.
Given:
Initial velocity, u = 79 ft/sec
Initial height, s = 4 ft
To find the maximum height, we need to determine the time it takes for the ball to reach its peak and then use that time to find the height.
Step 1: Find the time taken to reach the peak
Using the equation:
v = u + a*t
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time
At the peak, the final velocity will be zero (v = 0), so we can rearrange the equation to solve for t:
0 = 79 - 32*t
32*t = 79
t = 79/32
Step 2: Calculate the maximum height reached
Using the equation:
s = u*t + (1/2)*a*(t^2)
where s is the height, u is the initial velocity, a is the acceleration, and t is the time
Plugging in the values:
s = 79*(79/32) + (1/2)*(-32)*(79/32)^2
s = 79*(79/32) - 16*(79/32)^2
s ≈ 250.32 ft
Therefore, the ball will reach a height of approximately 250.32 feet.
To find the maximum height reached by the ball, we need to determine the time it takes for the ball to reach its highest point.
We know that the initial velocity (v₀) is 79 ft/sec and the acceleration due to gravity (a(t)) is -32 ft/sec². The negative sign indicates that the acceleration is directed downward.
Using the formula for velocity as a function of time, we have:
v(t) = v₀ + a(t)t
When the ball reaches its highest point, its velocity becomes zero, so we can set v(t) to zero and solve for the time it takes to reach this point. Thus, our equation becomes:
0 = 79 - 32t
Rearranging the equation, we have:
32t = 79
t = 79/32 ≈ 2.47 seconds
Now that we have determined the time it takes for the ball to reach its highest point, we can calculate the maximum height by substituting this time (t) into the equation for position as a function of time.
Using the formula for position as a function of time, we have:
s(t) = s₀ + v₀t + 0.5a(t)t²
Since the initial position (s₀) is 4 feet, the initial velocity (v₀t) is 79 ft/sec, and the acceleration due to gravity (a(t)) is -32 ft/sec², the equation becomes:
s(t) = 4 + 79t - 0.5(32)t²
Substituting the value of t ≈ 2.47 seconds into the equation, we can calculate the maximum height reached by the ball:
s(2.47) = 4 + 79(2.47) - 0.5(32)(2.47)² ≈ 160.59 feet
Therefore, the ball will reach a maximum height of approximately 160.59 feet.