Mr. Martin springs off a 5 meter high diving board with an upward velocity of 20 m/s.
A. Find Mr. Martin's height, velocity and acceleration after 0.5 seconds.
B. Determine when Mr. Martin will hit the water.
C. What is Mr. Martin's maximum height off the board?
D. What is Mr. Martin's maximum velocity when he hits the water?
E. What is Mr. Martin's velocity when he reaches half of his maximum height?
Danville is the name of a town in Illinois. Unless you have a class just to study that town, it is not your School Subject. If you want a tutor skilled in teaching physics to see your question, put the School Subject in the School Subject box.
According to your data, the height would be given by
h =-4.9t^2 + 20t + 5
totally absurd numbers, as you will see when we "solve" this problem
we have a downwards parabola and the t of the vertex is -20/-9.8
= appr 2.04 seconds
his maximum height is -4.9(2.04)^2 + 20(2.01) + 5
= 25.4 m , wow!! , looks like Superman has taken up diving.
When Clark hits the water, h = 0
-4.9t^2 + 20t + 5 = 0
Use the quadratic formula, reject the negative, and realize that he must have been in the air for over 4 seconds.
Did somebody just randomly make up these numbers? Surely no textbook would have that question, or does it??
To solve the given problem, we can use the equations of motion. These equations describe the motion of an object under constant acceleration. In this case, the acceleration due to gravity will be considered as constant throughout the motion.
A. To find Mr. Martin's height, velocity, and acceleration after 0.5 seconds, we need to use the following equations:
1. For height (h):
h = h0 + v0t + (1/2)at^2
where
h0 = initial height = 5 meters
v0 = initial velocity = 20 m/s (upward)
t = time = 0.5 seconds
a = acceleration = -9.8 m/s^2 (taking downwards as negative)
Plugging in the values, we have:
h = 5 + 20(0.5) + (1/2)(-9.8)(0.5)^2
= 5 + 10 - 1.225
= 13.775 meters (height after 0.5 seconds)
2. For velocity (v):
v = v0 + at
where
v0 = 20 m/s (upward)
t = 0.5 seconds
a = -9.8 m/s^2
Plugging in the values, we have:
v = 20 + (-9.8)(0.5)
= 20 - 4.9
= 15.1 m/s (velocity after 0.5 seconds)
3. For acceleration (a):
Since the acceleration due to gravity is constant, it remains -9.8 m/s^2 throughout the motion.
B. To determine when Mr. Martin will hit the water, we need to find the time it takes for him to reach the ground (height = 0). To do this, we can use the height equation mentioned earlier and solve for time (t):
h = h0 + v0t + (1/2)at^2
where
h0 = 5 meters
v0 = 20 m/s (upward)
t = ? (time till height = 0)
a = -9.8 m/s^2
Setting h = 0, we have:
0 = 5 + 20t + (1/2)(-9.8)t^2
Rearranging the equation and solving for t, we get:
-4.9t^2 + 20t + 5 = 0
Using the quadratic formula, we find two solutions for t: t = -0.113 seconds and t = 4.303 seconds.
The negative value is not meaningful in this context, so we discard it.
Therefore, Mr. Martin will hit the water after approximately 4.303 seconds.
C. Mr. Martin's maximum height off the board can be found using the same height equation. The maximum height occurs when the vertical velocity becomes 0 (at the peak of the motion). So we set v = 0 and solve for h:
v = v0 + at
0 = 20 + (-9.8)t
Solving for t, we get:
t = 20/9.8
= 2.04 seconds (time taken to reach maximum height)
Now, we can substitute this time value into the height equation:
h = 5 + 20(2.04) + (1/2)(-9.8)(2.04)^2
= 5 + 40.8 - 20.103
= 25.697 meters (maximum height off the board)
D. Mr. Martin's maximum velocity when he hits the water can be found using the equation of motion for velocity (v). At the point of impact, the height will be 0, so we can use this information to find the velocity:
v = v0 + at
0 = 20 + (-9.8)t
Solving for t, we get the same value as before: t ≈ 4.303 seconds.
Now we can substitute this time value into the velocity equation:
v = 20 + (-9.8)(4.303)
≈ -17.99 m/s (negative sign indicates downward direction)
Therefore, Mr. Martin's maximum velocity when he hits the water is approximately 17.99 m/s downward (towards the water).
E. To find Mr. Martin's velocity when he reaches half of his maximum height, we need to determine the corresponding time. Half of the maximum height would be h/2 = 25.697/2 ≈ 12.8485 meters.
Setting h = 12.8485 meters in the height equation:
12.8485 = 5 + 20t + (1/2)(-9.8)t^2
Rearranging and solving the equation, we find two solutions for t: t ≈ 0.645 seconds and t ≈ 3.395 seconds.
The time taken to reach half of the maximum height is approximately 0.645 seconds.
Now we can calculate Mr. Martin's velocity at this time using the velocity equation:
v = v0 + at
v = 20 + (-9.8)(0.645)
≈ 13.04 m/s (upward)
So, Mr. Martin's velocity when he reaches half of his maximum height is approximately 13.04 m/s (upward).