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?
v = Vi - 9.81 t
h = Hi + Vi t - 4.9 t^2
A.
after .5 seconds:
v = 20 - 9.81(.5)
h = 5 + 20 (.5) - 4.9 (.25)
B.
0 = 5 + 20 t - 4.9 t^2
solve quadratic equation for t when h is zero
C.
at top v = 0
0 = 20 - 4.9 t at top
use t at top for height
h = 5 + 20 t - 4.9 t^2
D.
v = 20 - 8.81 t
(it better be negative using the t from part B )
E.
use (1/2) the part C answer (I guess, it does not say off the water or off the board) for h in
h = 5 + 20 t - 4.9 t^2
get t then use it in
v = 20 - 9.81 t
of course there may be 2 answers for t, on the way up and on the way down.
By the way 20 meters per second is quite a spring. Are you sure it is not 2 ?
say you jump with a speed up of 20 m/s from the ground
at top. I will call g = 10 instead of 9.81
0 = 20 - 10 t
t = 2 seconds upward
h = 5 t^2 = 5(4) = 20 meters
that is like 20 yards
60 feet !!!
good grief.
To find the answers to these questions, we need to apply the principles of motion under gravity using the equations of motion. Here's how we can solve each of these questions:
A. Find Mr. Martin's height, velocity, and acceleration after 0.5 seconds:
1. We know that the initial velocity, u = 20 m/s and the acceleration due to gravity, g = 9.8 m/s² (assuming no air resistance).
2. Using the equation of motion, h = ut + (1/2)gt², where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
3. Plugging in the values, we have h = (20)(0.5) + (1/2)(9.8)(0.5)² = 10 + 1.225 = 11.225 meters.
To find the velocity after 0.5 seconds:
1. We can use the equation of motion, v = u + gt, where v is the final velocity.
2. Plugging in the values, we have v = 20 + (9.8)(0.5) = 20 + 4.9 = 24.9 m/s.
To find the acceleration after 0.5 seconds:
1. The acceleration remains constant at -9.8 m/s² (negative since it acts in the opposite direction of the initial velocity).
Therefore, after 0.5 seconds, Mr. Martin's height is 11.225 meters, velocity is 24.9 m/s, and acceleration is -9.8 m/s².
B. Determine when Mr. Martin will hit the water:
1. We can determine the time it takes for Mr. Martin to hit the water using the equation h = ut + (1/2)gt², where h is the height and u is the initial velocity.
2. Since the height of the water is 0 meters, we solve, 0 = (20)t + (1/2)(9.8)t².
3. Simplifying the equation, we get t² + 4t = 0.
4. Factoring out t, we have t(t + 4) = 0.
5. The solutions are t = 0 (which is the initial time) and t = -4 (which is not relevant in this context).
6. Therefore, Mr. Martin will hit the water after 0 seconds, which means he hits the water instantly when he jumps.
C. What is Mr. Martin's maximum height off the board:
1. Mr. Martin's maximum height will occur when his velocity becomes zero momentarily (at the highest point).
2. Using the equation v = u + gt, we can find the time it takes for his velocity to become zero.
3. Setting v = 0, we have 0 = 20 + (9.8)t.
4. Solving for t, we get t = -20 / 9.8 ≈ -2.04 seconds.
5. Since time cannot be negative in this case, we discard this solution.
6. Therefore, Mr. Martin's maximum height off the board is not attainable in this scenario. He reaches a height of 11.225 meters after 0.5 seconds.
D. What is Mr. Martin's maximum velocity when he hits the water:
1. Since Mr. Martin hits the water instantly, there is no time for his velocity to increase significantly due to gravity.
2. Therefore, his maximum velocity when he hits the water is approximately equal to his initial upward velocity of 20 m/s.
E. What is Mr. Martin's velocity when he reaches half of his maximum height:
1. Half of his maximum height would be approximately 5.6125 meters (half of 11.225).
2. To find the time it takes for Mr. Martin to reach this height, we can use the equation h = ut + (1/2)gt².
3. Setting h = 5.6125 and u = 20, we have 5.6125 = (20)t + (1/2)(9.8)t².
4. Simplifying the equation, we get 4.9t² + 20t - 5.6125 = 0.
5. Solving the quadratic equation, we find two values for t: t ≈ -0.82 seconds and t ≈ 0.54 seconds.
6. Since time cannot be negative in this case, we discard the solution t ≈ -0.82 seconds.
7. Therefore, Mr. Martin's velocity when he reaches half of his maximum height is approximately 20 m/s after 0.54 seconds.