John stand at the top of a 30m building. He throws a stone upwards with a speed of 5m/s. Calculate. (1)how long the stone is in the air. (2)the maxi
mum height the stone reaches the ground?
Start by solving for the height the stone reaches at the top of its trajectory. You can use the kinematic equation Vf^2=Vi^2+2gd where g is the acceleration due to gravity (I used 9.8) and d is the displacement. Knowing that the velocity in the y direction is 0 m/s when the rock is at the max y displacement, you can solve for d. Don't forget that g is negative. This will be the height the rock travels ABOVE the building. It comes out to be something like 1.28 m, so add 30 m and you have part 2.
For part two you should use the kinematic that is d=ViT+0.5GT^2. You'll have to use the quadratic equation to solve for t, but that's OK. Use the answer to part two (add the 30 m) for d and the 5 m/s for Vi. The positive solution is your answer. I got about 2 seconds.
Hope this helps!
To calculate the time the stone is in the air, we can use the equation of motion: displacement = initial velocity × time + (1/2) × acceleration × time².
Since the stone's initial velocity is 5m/s and it moves upwards, the acceleration due to gravity should be negative (assuming upward is positive). Therefore, the equation becomes:
-30m = 5m/s × t + (1/2) × (-9.8m/s²) × t²
Simplifying the equation:
-30m = 5t - 4.9t²
Rearranging the equation:
4.9t² - 5t - 30 = 0
Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
For this equation:
a = 4.9
b = -5
c = -30
Substituting these values:
t = (-(-5) ± √((-5)² - 4 × 4.9 × -30)) / (2 × 4.9)
Simplifying:
t = (5 ± √(25 + 588)) / 9.8
t = (5 ± √613) / 9.8
This gives us two possible values for t, one positive and one negative. Since time cannot be negative in this context, we consider the positive value:
t ≈ 1.93 seconds
So, the stone is in the air for approximately 1.93 seconds.
To calculate the maximum height reached by the stone, we can use the equation: final velocity² = initial velocity² + 2 × acceleration × displacement.
Since the stone is thrown upwards with an initial velocity of 5m/s, the final velocity at the maximum height will be 0m/s. Therefore, the equation becomes:
0m/s² = (5m/s)² + 2 × (-9.8m/s²) × displacement
Simplifying the equation:
0 = 25 - 19.6 × displacement
Rearranging the equation:
Displacement = 25 / 19.6
Displacement ≈ 1.28 meters
Thus, the maximum height reached by the stone is approximately 1.28 meters above its starting point on the ground.