A particle travels between two parallel ver-
tical walls separated by 39 m. It moves to-
ward the opposing wall at a constant rate of
8.8 m/s. It hits the opposite wall at the same
height.
The acceleration of gravity is 9.8 m/s2 .
39 m
9.8 m/s2
8.8 m/s
a) What will be its speed when it hits the
opposing wall?
Answer in units of m/s
To find the speed of the particle when it hits the opposing wall, you can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity (8.8 m/s)
a = acceleration due to gravity (-9.8 m/s^2, negative because it is acting opposite to the direction of motion)
s = displacement (39 m)
Plugging in the values:
v^2 = (8.8 m/s)^2 + 2(-9.8 m/s^2)(39 m)
v^2 = 77.44 m^2/s^2 + (-764.4 m^2/s^2)
v^2 = -686.96 m^2/s^2
Since speed cannot be negative, we take the positive square root:
v = √(-686.96) m/s
v ≈ 26.2 m/s (rounded to one decimal place)
Therefore, the speed of the particle when it hits the opposing wall is approximately 26.2 m/s.
To find the speed of a particle when it hits the opposing wall, we can use the concept of uniform acceleration.
Given:
Distance traveled (d) = 39 m
Acceleration due to gravity (g) = 9.8 m/s^2
Initial velocity (u) = 8.8 m/s
We can use the equation of motion:
v^2 = u^2 + 2ad
Where:
v is the final velocity
u is the initial velocity
a is the acceleration
d is the distance traveled
In this case, the acceleration is due to gravity, which is acting vertically downwards. So, the acceleration will be -9.8 m/s^2 (negative because it is acting in the opposite direction to the particle's motion).
Substituting the given values into the equation, we get:
v^2 = (8.8 m/s)^2 + 2(-9.8 m/s^2)(39 m)
v^2 = 77.44 m^2/s^2 + (-764.40 m^2/s^2)
v^2 = -686.96 m^2/s^2
Since we're interested in the speed, which is a scalar quantity, we take the positive square root of the result:
v = √(-686.96 m^2/s^2)
v ≈ 26.20 m/s
Therefore, the speed of the particle when it hits the opposing wall is approximately 26.20 m/s.