Refer to the diagram shown below.
h = height of the girl above water when she lets go of the rope.
The launch velocity is 22.5 m/s at 35° to the horizontal. Therefore the vertical component of the velocity is
v = 22.5 sin(35°) = 12.9055 m/s.
The time of flight is t = 1.10 s before the girl hits the surface of the water at a height of -h.
Therefore
-h = (12.9055 m/s)*(1.10 s) - (1/2)*(9.8 m/s²)*(1.10 s)²
-h = 8.267 m
= 8.3 m (nearest tenth)
Answer:
When the girl let go of the rope, she was about 8.3 m above the surface of the water.
Answer:
10.2 metres
Explanation:
Given that a ball is projected at an initial speed of 20.0 meter per second making an angle of 45.0 with horizontal. What is the maximum height it will reach?
Solution
To get the maximum height, let us use the formula
V^2 = U^2 sin^2ø - 2gH
At maximum height V = 0
U^2 sin^2ø = 2gH
Substitute all the parameters into the formula
20^2 ( sin 45 )^2 = 2 × 9.8 × H
400 × 0.5 = 19.6 H
Make H the subject of formula
H = 200 / 19.6
H = 10.204 metres.
Therefore, the maximum height reached by the projected ball is 10.2 metres.
Answer:
The answer is 24cm
Explanation:
This problem bothers on the curved mirrors, a concave type
Given data
Object height h= 5cm
Object distance = 12cm
Focal length f=24cm
Let the image distance be v=?
Applying the formula we have
1/v +1/u= 1/f
Substituting our given data
1/v+1/12=1/24
1/v=1/24-1/12
1/v=1-2/24
1/v=-1/24
v= - 24cm
This implies that the image is on the same side as the object and it is real
The answer would be:
A, a space station orbiting Earth.