Drum major tosses the baton into the air from height of 6feet and initial velocity of 45 ft/sec. What is max height? We are to use ht=h+vt-1/2at^2
V = Vo + g*t
t = (V-Vo)/g = (0-45)/-32 = 1.41 s.
ht. = 6 + 45*1.41 - 0.5*32*(1.41)^2 = 37.64 Ft.
To find the maximum height reached by the baton, we can use the equation of motion:
ht = h + vt - 1/2at^2
where:
ht = final height (maximum height)
h = initial height (6 feet)
v = initial velocity (45 ft/sec)
t = time taken to reach the maximum height
a = acceleration due to gravity (32.2 ft/sec^2, assuming upward as positive direction)
Now, let's solve the equation step-by-step:
Step 1: Plug in the given values into the equation:
ht = 6 + 45t - 1/2(32.2)t^2
Step 2: At the maximum height, the velocity becomes 0. So, we need to find the time it takes for the velocity to become zero. To do this, we set the equation for velocity equal to zero and solve for t:
0 = v - at
0 = 45 - 32.2t
Step 3: Solve the equation:
32.2t = 45
t = 45/32.2
t ≈ 1.398 seconds
Step 4: Substitute the value of t into the equation for ht:
ht = 6 + 45(1.398) - 1/2(32.2)(1.398)^2
Step 5: Perform the calculations:
ht ≈ 6 + 62.91 - 31.44
ht ≈ 37.47 feet
Therefore, the maximum height reached by the baton is approximately 37.47 feet.
To find the maximum height reached by the baton, you can use the equation of motion:
ht = h + vt - 1/2 * at^2
Where:
ht = maximum height (what we're trying to find)
h = initial height (given as 6 feet)
vt = initial vertical velocity (given as 45 ft/sec)
a = acceleration due to gravity (-32 ft/sec^2, as the object is moving upward against the force of gravity)
t = time taken for the object to reach maximum height
Step 1: Find the time taken to reach maximum height (t):
Since the baton is thrown vertically upwards, its final velocity at maximum height will be zero. So we can use the equation:
vf = vi + at
0 = 45 - 32t
Rearranging the equation, we get:
32t = 45
t = 45/32 = 1.40625 seconds
Step 2: Substitute the values into the equation to find the maximum height (ht):
ht = h + vt - 1/2 * at^2
= 6 + (45 * 1.40625) - 1/2 * (-32) * (1.40625)^2
Simplifying further, we have:
ht = 6 + 63.28125 + 17.5
ht = 86.78125 feet
Therefore, the maximum height reached by the baton is approximately 86.78125 feet.