A stone thrown from the top of a building which is 30m is given an initial velocity 20m/s verticaly upward.the stone misses the edge of the building on its way down.
A,what is the velocity of the stone after 1second,n after 3seconds.
B,find the maximum height reached by the stone .
C,how long does it take the stone to reach the ground.
Sorry -- but none of the Jiskha tutors are expert in your stated school subject: Uniswa.
A. 2.04s
B. 20.41m
C. 4.08s
A) To find the velocity of the stone after 1 second, we can use the equation for vertical motion:
v = u + at
Given:
Initial velocity, u = 20 m/s (upward)
Acceleration, a = -9.8 m/s^2 (due to gravity, in the downward direction)
After 1 second:
Time, t = 1 second
Using the equation, we can substitute the values:
v = 20 + (-9.8)(1)
v = 20 - 9.8
v ≈ 10.2 m/s (upward)
After 3 seconds:
Time, t = 3 seconds
Using the equation again:
v = 20 + (-9.8)(3)
v = 20 - 29.4
v ≈ -9.4 m/s (upward)
Note: The negative sign indicates that the stone is moving downward.
B) To find the maximum height reached by the stone, we can use the equation:
v^2 = u^2 + 2as
At the maximum height, the final velocity is 0 m/s.
Given:
Initial velocity, u = 20 m/s (upward)
Acceleration, a = -9.8 m/s^2 (due to gravity, in the downward direction)
Final velocity, v = 0 m/s
Using the equation, we can substitute the values:
0 = 20^2 + 2(-9.8)s
0 = 400 - 19.6s
19.6s = 400
s ≈ 20.41 m
Therefore, the maximum height reached by the stone is approximately 20.41 meters.
C) To find how long it takes for the stone to reach the ground, we can use the equation:
s = ut + (1/2)at^2
At the moment it hits the ground, s = 30 m (height of the building), u = 20 m/s (upward), and a = -9.8 m/s^2 (due to gravity, in the downward direction).
Let's solve for the time, t:
30 = 20t + (1/2)(-9.8)t^2
0 = -4.9t^2 + 20t - 30
Simplifying the equation, we get a quadratic equation:
4.9t^2 - 20t + 30 = 0
We can solve this equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values into the formula:
t = (-(-20) ± √((-20)^2 - 4(4.9)(30))) / (2(4.9))
t = (20 ± √(400 - 588)) / 9.8
t = (20 ± √(-188)) / 9.8
Since we cannot take the square root of a negative number in this case, it means that the stone will not reach the ground within the given conditions.
To answer these questions, we can use the equations of motion for an object in free fall.
The equations of motion for an object in free fall are:
1. Distance (s) = Initial velocity (u) × Time (t) + 0.5 × Acceleration (a) × Time (t)^2
2. Final velocity (v) = Initial velocity (u) + Acceleration (a) × Time (t)
First, let's calculate the acceleration due to gravity. The acceleration due to gravity is approximately 9.8 m/s^2, and since the stone is thrown vertically upward, we will take it as negative (opposite to the direction of motion).
A) Velocity after 1 second:
Using the second equation of motion, we can calculate the final velocity after 1 second.
Initial velocity (u) = 20 m/s (upward)
Acceleration (a) = -9.8 m/s^2
Time (t) = 1 second
Using the equation: v = u + a × t
v = 20 m/s - (9.8 m/s^2 × 1 s)
v = 20 m/s - 9.8 m/s
v = 10.2 m/s
The velocity of the stone after 1 second is 10.2 m/s (upward).
Velocity after 3 seconds:
Using the same equation, we can calculate the final velocity after 3 seconds.
Initial velocity (u) = 20 m/s (upward)
Acceleration (a) = -9.8 m/s^2
Time (t) = 3 seconds
Using the equation: v = u + a × t
v = 20 m/s - (9.8 m/s^2 × 3 s)
v = 20 m/s - 29.4 m/s
v = -9.4 m/s
The velocity of the stone after 3 seconds is -9.4 m/s (downward).
B) Maximum height reached by the stone:
To find the maximum height reached by the stone, we need to find the time it takes for the stone to reach the highest point. At the highest point, the velocity becomes zero.
Using the equation: v = u + a × t
0 = 20 m/s - 9.8 m/s^2 × t
Simplifying the equation, we get:
9.8 m/s^2 × t = 20 m/s
t = 20 m/s ÷ 9.8 m/s^2
t ≈ 2.04 seconds
Now, we can use the time obtained to find the maximum height using the first equation of motion.
Using the equation: s = u × t + 0.5 × a × t^2
s = 20 m/s × 2.04 s + 0.5 × -9.8 m/s^2 × (2.04 s)^2
Simplifying the equation, we get:
s = 20.4 m - 20. %m/s^2 × (4.1616 s)^2
s ≈ 20.4 m - 20.4 m
s ≈ 0 m
The maximum height reached by the stone is approximately 0 meters, which means the stone does not go any higher than the initial height of the building.
C) Time taken to reach the ground:
To find the time taken to reach the ground, we can use the equation for distance:
Using the equation: s = u × t + 0.5 × a × t^2
where, s = -30 m (negative because the stone is moving downward)
u = 20 m/s (initial upward velocity)
a = -9.8 m/s^2 (acceleration due to gravity)
t = ?
Rearranging the equation, we get:
0.5 × -9.8 m/s^2 × t^2 + 20 m/s × t - 30 m = 0
Solving this quadratic equation, we find two roots. However, we can neglect the negative root since we are looking for a positive time.
Using the quadratic formula, we get:
t = (-20 m/s ± √(20 m/s)^2 - 4 × 0.5 × -9.8 m/s^2 × -30 m) / (2 × 0.5 × -9.8 m/s^2)
Simplifying the equation, we get:
t = (20 m/s ± √(400 m^2/s^2 + 588 m^2/s^2)) / -9.8 m/s^2
t ≈ (20 m/s ± √988 m^2/s^2) / -9.8 m/s^2
t ≈ (20 m/s ± 31.44 m/s) / -9.8 m/s^2
The positive root gives us the time taken to reach the ground:
t ≈ (20 m/s + 31.44 m/s) / -9.8 m/s^2
t ≈ 3.15 s
Therefore, it takes approximately 3.15 seconds for the stone to reach the ground.