Question

a stone is thrown with an initial speed of 13.7 m/s at an angle of 57 degrees above the horizontal from the top of a 32 m tall building. if air resistance is negligible, then what is the magnitude of the horizontal component of velocity when the rock strikes the ground?

Answer 1

To find the magnitude of the horizontal component of velocity when the rock strikes the ground, we first need to calculate the time it takes for the stone to hit the ground. We can use the equation h = (1/2) * g * t^2. Now, we can find the horizontal component of velocity using the equation v = d / t.

Step-by-step explanation:

To find the magnitude of the horizontal component of velocity when the rock strikes the ground, we first need to calculate the time it takes for the stone to hit the ground. We can use the equation:

h = (1/2) * g * t^2

Where h is the height of the building, g is the acceleration due to gravity (9.8 m/s^2), and t is the time of flight. Rearranging the equation, we get:

t = sqrt((2h) / g)

= sqrt((2 * 32) / 9.8)

= 2 s

Now, we can find the horizontal component of velocity using the equation:

v = d / t

= 100 / 2

= 50 m/s

Therefore, the magnitude of the horizontal component of velocity when the rock strikes the ground is 50 m/s.