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Help please! It's a brain Twister!?
A box left the window at a horizontal speed of 20 metres per second, and the window was at a height of 45 metres, and the ground outside of the tower was perfectly horizontal, and the box's mass was 6.2 kg. Assume gravitational acceleration of 9.8 metres per second squared.
How much horizontal distance, in centimetres, will the box travel from the tower before hitting the ground? Assume no additional forces (such as air resistance, insect collisions, etc) are acting on the box and round to the nearest centimeter. For example, if the box lands 283.8 metres away, you'd submit the answer "28380"
2 Answers
- TychaBraheLv 71 decade agoFavorite Answer
The object has an initial vertical velocity of 0 m/s. It falls 45 m accelerating at 9.8 m/s^2
delta-s = (1/2) * a * t^2
45 m = (1/2) * 9.8 m/s^2 * t^2
t^2 = 45 m / ((1/2) * 9.8 m/s^2)
t^2 = 9.184
t = 3.030 s
So it takes the object 3.030 s to fall.
How far will its horizontal speed of 20 m/s allow it travel in 3.03 s?
s = vt
s = 20 m/s * 3.030 s
s = 60.61 m
- Anonymous1 decade ago
This is a projectile motion problem in two dimensions. First you need to find the time it takes for the box to hit the ground, which can be found using the kinematic equation d = v1t + 1/2*a*t^2. Here a is the vertical acceleration, equal to g. Solve the quadratic for t and you have the time.
There will be no acceleration in the x direction assuming no air resistance. Thus now that you know the flight time, you only have to use the equation d = v * t with the initial speed of 20m/s to get the horizontal distance travelled.
