An object is being thrown upward at 50m How long does it take the object to reach maximum height? What is the maximum height of the object? How much time does it take for the audit to return to its original position? What is the velocity of the object when it hits the ground? How much time Total does it take for the advent to hit the ground?
A motion question so think about the equations of motion:
v = u + at
v^2 = u^2 + 2aS
S = ut - (1/2)*a*t^2
u = initial vel
v = final vel
t = time
S = displacement
a = acceleration which in this case is gravity so g = 9.8 m/s^2
Don't forget that when throwing up gravity is against you so a = -g but when it's going downwards gravity is helping it so a = g then.
Write down what you know for the first part and see which equation you can use to find the time (t)
edit:
S = ut + (1/2)*a*t^2
Sorry about that. I was thinking of the upward throw and the - sneaked in.
To answer these questions, we can use the equations of motion for objects in free fall.
1. How long does it take the object to reach maximum height?
To find the time it takes for the object to reach maximum height, we can use the equation:
t = (v_final - v_initial) / (-g)
where t is the time, v_final is the final velocity, v_initial is the initial velocity, and g is the acceleration due to gravity. In this case, the object is being thrown upward, so the initial velocity is positive 50 m/s, and the final velocity at maximum height is 0 m/s as it momentarily stops before beginning to fall. The acceleration due to gravity is negative 9.8 m/s² as it acts downward. Plugging the values into the equation:
t = (0 - 50) / (-9.8) = 5.102 seconds (approx.)
Therefore, it takes approximately 5.102 seconds for the object to reach maximum height.
2. What is the maximum height of the object?
To find the maximum height, we can use the equation:
h = v_initial * t + (1/2) * g * t²
where h is the height. Plugging in the values:
h = 50 * 5.102 + (1/2) * (-9.8) * (5.102)² = 127.55 meters (approx.)
Therefore, the maximum height of the object is approximately 127.55 meters.
3. How much time does it take for the object to return to its original position?
Since the object was thrown upward and will eventually fall back to its original position, the total time it takes can be found by doubling the time it took to reach the maximum height:
Total time = 2 * 5.102 = 10.204 seconds (approx.)
Therefore, it takes approximately 10.204 seconds for the object to return to its original position.
4. What is the velocity of the object when it hits the ground?
To find the velocity of the object when it hits the ground, we can use the equation:
v_final = v_initial + g * t
Plugging in the values:
v_final = 50 + (-9.8) * 10.204 = -99.996 m/s (approx.)
Note that the negative sign indicates the velocity is downward.
Therefore, the velocity of the object when it hits the ground is approximately -99.996 m/s.
5. How much total time does it take for the object to hit the ground?
The total time it takes for the object to hit the ground can be found by doubling the time it took to reach the maximum height and adding it to the time it took to reach the maximum height:
Total time = 5.102 + 10.204 = 15.306 seconds (approx.)
Therefore, it takes approximately 15.306 seconds for the object to hit the ground.