Answer:
30.63 m
Explanation:
From the question given above, the following data were obtained:
Total time (T) spent by the ball in air = 5 s
Maximum height (h) =.?
Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:
Total time (T) spent by the ball in air = 5 s
Time (t) taken to reach the maximum height =.?
T = 2t
5 = 2t
Divide both side by 2
t = 5/2
t = 2.5 s
Thus, the time (t) taken to reach the maximum height is 2.5 s
Finally, we shall determine the maximum height reached by the ball as follow:
Time (t) taken to reach the maximum height = 2.5 s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =.?
h = ½gt²
h = ½ × 9.8 × 2.5²
h = 4.9 × 6.25
h = 30.625 ≈ 30.63 m
Therefore, the maximum height reached by the cannon ball is 30.63 m
Answer:
Option C is the untrue statement.
Answer:
400m
Explanation:
Brainliest? :))
Let your initial displacement from your home to the store be
Dd
>
1 and your displacement from the store to your friend’s house
be Dd
>
2.
Given: Dd
>
1 = 200 m [N]; Dd
>
2 = 600 m [S]
Required: Dd
>
T
Analysis: Dd
>
T 5 Dd
>
1 1 Dd
>
2
Solution: Figure 6 shows the given vectors, with the tip of Dd
>
1
joined to the tail of Dd
>
2. The resultant vector Dd
>
T is drawn in red,
from the tail of Dd
>
1 to the tip of Dd
>
2. The direction of Dd
>
T is [S].
Dd
>
T measures 4 cm in length in Figure 6, so using the scale of
1 cm : 100 m, the actual magnitude of Dd
>
T is 400 m.
Statement: Relative to your starting point at your home, your
total displacement is 400 m [S].
Answer:
The third stage of parturition is called " after-birth".
Explanation:
To answer this problem, we will use the equations of motions.
Part (a):
For the ball to start falling back to the ground, it has to reach its highest position where its final velocity will be zero.
The equation that we will use here is:
v = u + at where
v is the final velocity = 0 m/sec
u is the initial velocity = 160 m/sec
a is acceleration due to gravity = -9.8 m/sec^2 (the negative sign is because the ball is moving upwards, thus, its moving against gravity)
t is the time that we want to find.
Substitute in the equation to get the time as follows:
v = u + at
0 = 160 - 9.8t
9.8t = 160
t = 160/9.8 = 16.3265 sec
Therefore, the ball would take 16.3265 seconds before it starts falling back to the ground
Part (b):
First, we will get the total distance traveled by the ball as follows:
s = 0.5 (u+v)*t
s = 0.5(160+0)*16.3265
s = 1306.12 meters
The equation that we will use to solve this part is:
v^2 = u^2 + 2as where
v is the final velocity we want to calculate
u is the initial velocity of falling = 0 m/sec (ball starting falling when it reached the highest position, So, the final velocity in part a became the initial velocity here)
a is acceleration due to gravity = 9.8 m/sec^2 (positive as ball is moving downwards)
s is the distance covered = 1306.12 meters
Substitute in the above equation to get the final velocity as follows:
v^2 = u^2 + 2as
v^2 = (0)^2 + 2(9.8)(1306.12)
v^2 = 25599.952 m^2/sec^2
v = 159.99985 m/sec
Therefore, the velocity of the ball would be 159.99985 m/sec when it hits the ground.
