Answer: 16.3 seconds
Explanation: Given that the
Initial velocity U = 80 ft/s
Let's first calculate the maximum height reached by using third equation of motion.
V^2 = U^2 - 2gH
Where V = final velocity and H = maximum height.
Since the toy is moving against the gravity, g will be negative.
At maximum height, V = 0
0 = 80^2 - 2 × 9.81 × H
6400 = 19.62H
H = 6400/19.62
H = 326.2
Let's us second equation of motion to find time.
H = Ut - 1/2gt^2
Let assume that the ball is dropped from the maximum height. Then,
U = 0. The equation will be reduced to
H = 1/2gt^2
326.2 = 1/2 × 9.81 × t^2
326.2 = 4.905t^2
t^2 = 326.2/4.905
t = sqrt( 66.5 )
t = 8.15 seconds
The time it will take for the rocket to return to ground level will be 2t.
That is, 2 × 8.15 = 16.3 seconds
Velocity is speed and the direction of the speed. Acceleration is the rate at which Velocity is changing, and the direction in which it's changing.
Answer:
Hyperopia
Explanation:
In hyperopia ,people face difficulties to see close up object , but can see object easily which are at a distance.
The main reason of hyperopia is our eyeball.When our eyeball become too short , then light focus behind the retina. Sowe will face problem to see near object but we can see distance object easily. Hyperopia is the opposite of nearsightedness. Hyperopia can be corrected by using contact lenses.
Answer:
9/25
Explanation:
Distance covered in the first 5 seconds:
Δx = v₀ t + ½ at²
Δx₀₋₅ = (0) (5) + ½ a (5)²
Δx₀₋₅ = 25a/2
Distance covered in the first 4 seconds:
Δx = v₀ t + ½ at²
Δx₀₋₄ = (0) (4) + ½ a (4)²
Δx₀₋₄ = 8a
So the distance covered during the 5th second is:
Δx₅ = 25a/2 − 8a
Δx₅ = 9a/2
So the ratio of the distance covered during the 5th second to the distance covered in the first 5 seconds is:
Δx₅ / Δx₀₋₅
(9a/2) / (25a/2)
9/25
