Identical cannonballs are fired with the same force, one each from four cannons having respective bore lengths of 1.0 meter, 2.0
2 answers:
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Answer:
x = 240 m
Explanation:
This is a kinematics exercise
Let's fix our frame of reference on car A
x = x₀ₐ+ v₀ₐ t + ½ aₐ t²
the initial position of car a is zero
x = 0 + v₀ₐ t + ½ 0.8 t²
for car B
x = x_{ob} + v_{ob} t - ½ a_b t²
car B's starting position is 30 m
x = 30 + v_{ob} t - ½ 0.4 t²
at the point where they meet, the position of the two vehicles is the same
0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²
let's reduce the speeds to the SI system
v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s
v_{ob} = 23.4 km / h = 6.5 m / s
4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²
0.2 t² - 2.5 t - 30 = 0
t² - 12.5 t - 150 = 0
we solve the quadratic equation
t =
t =
t₁ = 20 s
t₂ = -7.5 s
time must be a positive quantity so the correct result is t = 20 s
let's look for the distance
x = 4 t + ½ 0.8 t²
x = 4 20 + ½ 0.8 20²
x = 240 m
Answer:
The person can jump 48 m on the Moon
Explanation:
The question parameters are;
The maximum long jump distance of a person on Earth, = 3 m
The acceleration due to gravity on the Moon = 1 ÷ 16 of that on Earth
The distance the person can jump on the Moon is given as follows;
A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile
The horizontal range, , of a projectile motion is found by using the following formula
Where;
g = The acceleration due to gravity = 9.8 m/s²
Therefore, we have;
u² = 3 m × 9.8 m/s² = 29.4 m²/s²
Therefore, on the Moon, we have;
The acceleration due to gravity on the Moon, = 1/16 × g
∴ = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²
The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is ≈ 48 m