An objectis thrown upward from the edge of a tall building with a velocity of 10m/s. Where will the object be 3s after it is thrown? Take g = 10m/s^2
15m above the top of the building
30m below the top of the building
15m below the top of the building
30m above the building
d = Vo*t + 0.5g*t^2,
d = 10*3 - 5*3^2 = -15m. = 15m below the top of building.
What is the gravitation field strength of a height above the surface of the earth? R is the radius of the earth. Ans a, pls correct
To determine where the object will be 3 seconds after it is thrown, we can use the kinematic equation:
h = initial_height + initial_velocity * time - 0.5 * acceleration * time^2
Given:
Initial velocity (u) = 10 m/s (upward)
Time (t) = 3 s
Acceleration due to gravity (g) = 10 m/s^2 (downward)
We need to find the position or height (h) of the object 3 seconds later.
Let's substitute the values into the equation:
h = 0 + (10 m/s * 3 s) - 0.5 * (10 m/s^2) * (3 s)^2
Simplifying further,
h = 0 + 30 m - 0.5 * 10 m/s^2 * 9 s^2
h = 0 + 30 m - 0.5 * 10 m/s^2 * 81 s^2
h = 30 m - 0.5 * 10 m/s^2 * 81 s^2
h = 30 m - 0.5 * 10 m/s^2 * 81 s^2
h ≈ 30 m - 40.5 m
h ≈ -10.5 m
The position (height) of the object after 3 seconds is approximately 10.5 meters below the top of the building. Therefore, the correct answer is:
30m below the top of the building