A 2kg block rests at the edge of a platform that is 10m above ground level.The block launched horizontally from the edge of the platform with an initial speed of 3m/s. Air resistance is negligible.what is the time taken for the block to reach the ground?

0.5g*t^2 = 10.

4.9t^2 = 10,
t = 1.43 s.

vertical free fall equation ... h = -1/2 g t^2 + Vo t + Ho

0 = -1/2 * 9.8 * t^2 + 0 + 10 = -4.9 t^2 + 10

solve for t (quadratic formula)

the horizontal motion has no effect on the flight time

time to fall: t=sqrt(2h/g)

solve that for time.

To determine the time taken for the block to reach the ground, we can use the equations of motion.

Let's consider the vertical motion of the block. Since air resistance is negligible, the only force acting on the block in the vertical direction is the force due to gravity. Using the equation:

s = ut + (1/2)gt^2

where s is the vertical displacement, u is the initial vertical velocity, g is the acceleration due to gravity, and t is the time taken.

In this case, we know that the initial vertical velocity (u) is 0 m/s, as the block is launched horizontally. The vertical displacement (s) is equal to the height of the platform (10 m), and the acceleration due to gravity (g) is approximately 9.8 m/s^2.

Plugging the known values into the equation:

10 = 0 * t + (1/2) * 9.8 * t^2

Simplifying the equation:

5t^2 = 10

Dividing both sides by 5:

t^2 = 2

Taking the square root of both sides:

t = √2

Therefore, the time taken for the block to reach the ground is approximately √2 seconds.