Explanation:
∆x=300 m×2
∆t=1.5 s
v=∆x/∆t → v=2×300/1.5 = 400 m/s
Answer:
The answer is below
Explanation:
a) The initial velocity (u) = 24 m/s
We can solve this problem using the formula:
v² = u² - 2gh
where v = final velocity, g= acceleration due to gravity = 9.8 m/s², h = height.
At maximum height, the final velocity = 0 m/s
v² = u² - 2gh
0² = 24² - 2(9.8)h
2(9.8)h = 24²
2(9.8)h = 576
19.6h = 576
h = 29.4 m
b) The time taken to reach the maximum height is given as:
v = u - gt
0 = 24 - 9.8t
9.8t = 24
t = 2.45 s
The total time needed for the apple to return to its original position = 2t = 2 * 2.45 = 4.9 s
Answer:
Given:
m=1000kg
u= 16.7m/s
v=0m/s
F=8000N
Required:
s=?
Solution:
F=m × a
8000N=1000kg × a
a=8m/s^2
Since it decelerate a= -8m/s^2
v^2 = u^2 + 2as
s=v^2 - u^2 / 2a
s= 0 - (16.7m/s)^2 / 2 × -8m/s^2
s= -278.89/-16
s= 17.43m
The car travels approximately 17.43m before it stops
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