A ball is thrown vertically up from the edge of a 150 m building with a veloicty of 10m/s. (A) how high did the ball rise? (b) how much time did it take fort he ball to reach this height ? (C) what is the velocity oft he ball as it strikest he ground ? (D) how long does it take for the ball to reacht he ground from its starting point (e) what is the position and velocity 1 sec and 3 sec after it is thrown ?
v = Vi - 9.81 t
v = 0 at top
so
t = 10/9.81 at top (part B)
Htop = 150 + 10 t - 4.9 t^2 (part A)
now simply drops from Htop to ground
Pe at top = m g Htop
Ke at bottom = (1/2) m v^2
so
v = sqrt (2 g Htop) (part C)
we have original t rising to top
now time falling
v = 0 + g Tfall
we know v from part C
so
Tfall = v/9.81
add that to our answer to part B to get total time in air. That is part D
Part E:
v = Vi - 9.81 t
h = Hi + Vi t - 4.9 t^2
Thank you very much
I don't have to add Tfall with the part b beacuse starting point is 150 and not in the air
That t top in part B is the time to slow down on the way up from the top of the building to the max height. 10/9.81 seconds.
THEN, it starts falling from that top point I called Htop. By the way it will pass the roof on the way down at 10 m/s down :)
To solve these questions, we will make use of the equations of motion for vertical motion under gravity. These equations are:
1. Displacement:
- For an object thrown vertically upwards: S = ut - (1/2)gt^2
- For an object falling downwards: S = ut + (1/2)gt^2
Where:
- S is the displacement (upwards or downwards).
- u is the initial velocity.
- t is the time.
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. Velocity:
- v = u - gt
Where:
- v is the final velocity.
- u is the initial velocity.
- g is the acceleration due to gravity.
Now, let's answer the given questions:
(A) How high did the ball rise?
To find the height the ball reached, we need to calculate the maximum height reached by the ball. At the maximum height, the velocity will be zero.
Using the velocity equation, we can find the time taken to reach the maximum height:
0 = 10 - 9.8t
Solving for t, we get t = 10/9.8 = 1.02 s.
Now, substituting this value of t into the displacement equation while taking the initial velocity as 10 m/s (since the ball is thrown upwards):
S = 10(1.02) - (1/2)(9.8)(1.02)^2
S = 10.2 - 5.04
S = 5.16 m
So, the ball reached a height of 5.16 meters.
(B) How much time did it take for the ball to reach this height?
We already calculated the time in the previous step. The ball took 1.02 seconds to reach the maximum height.
(C) What is the velocity of the ball as it strikes the ground?
The velocity of the ball just before striking the ground would be equal to the initial velocity, but in the opposite direction since it's falling.
Therefore, the velocity is -10 m/s.
(D) How long does it take for the ball to reach the ground from its starting point?
To find the time taken to reach the ground, we can use the displacement equation:
150 = 10t - (1/2)(9.8)t^2
Rearranging and solving the quadratic equation, we get t ≈ 10.2 s or t ≈ 0 s.
Since time cannot be negative, we discard the t ≈ 0 s solution.
So, it takes approximately 10.2 seconds for the ball to reach the ground.
(E) What is the position and velocity 1 second and 3 seconds after it is thrown?
To find the position and velocity at a given time, we can use the displacement and velocity equations.
For 1 second after it is thrown:
Position:
S = 10(1) - (1/2)(9.8)(1)^2
S = 10 - 4.9
S ≈ 5.1 m
Velocity:
v = 10 - 9.8(1)
v ≈ 0.2 m/s
For 3 seconds after it is thrown:
Position:
S = 10(3) - (1/2)(9.8)(3)^2
S = 30 - 44.1
S ≈ -14.1 m
Velocity:
v = 10 - 9.8(3)
v ≈ -19.4 m/s
So, 1 second after it is thrown, the ball is approximately 5.1 meters above its starting point with a velocity of 0.2 m/s. 3 seconds after it is thrown, the ball is approximately 14.1 meters below its starting point with a velocity of -19.4 m/s.