The height of the roof is <u>3.57m</u>
Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.
Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.
The length of the window s is given by,
The first drop is at the bottom and it takes 5t seconds to reach down.
The height of the roof h is the distance traveled by the first drop and is given by,
the height of the roof is 3.57 m
Answer:
Explanation:
The formula S=(at^2)/2 will be used during the entire explanation.
1. 4 = (2t^2)/2
t = 2 s.
V = at = 2 * 2 = 4 m/s
2. 8 = (2t^2)/2
t = 2.8 s
iii. 2s
iv. 2.8 - 2 = 0.8s
Hope you understand)
Answer:
Only at point a
Explanation:
If the velocity is greater than zero, than the object must be moving.
Answer:
1.) 1620 km/h^2
2.) 2.7 km
Explanation:
1.) Given that the car start from rest. The initial velocity U will be equal to zero. That is,
U = 0.
Final velocity V = 54 km/h
Time t = 2 minute = 2/60 = 1/30 hour
Acceleration a will be change in velocity per time taken. That is,
a = ( V - U )/ t
Substitute V, U and t into the formula
a = 54 ÷ 1/30
a = 54 × 30 = 1620 km/h ^2
2.) Distance travelled S by the car during the time can be calculated by using the 2nd equation of motion.
S = Ut + 1/2at^2
Substitute all the parameters into the formula
S = 54 × 1/30 + 1/2 × 1620 × (1/30)^2
S = 54/30 + 810 × 1/900
S = 54/30 + 810/900
S = (1620+810)/900
S = 2430/900
S = 2.7 km.
Therefore, distance travelled by the car during this time is 2.7 km
