A baseball is thrown horizontally off a 40-m high cliff with a speed of 16.3 m/s. How much time in seconds will it take for the baseball to hit the ground?
h = 0.5g*t^2 = 40 m.
4.9t^2 = 40
t^2 = 8.16
t = 2.86 s.
To solve this problem, we can use the kinematic equation for vertical motion:
y = y0 + v0yt - (1/2)gt^2
Where:
- y is the vertical position (40 m, since the baseball is thrown off a 40-m high cliff)
- y0 is the initial vertical position (0 m, as the base of the cliff is the reference point)
- v0y is the initial vertical velocity (0 m/s, as the baseball is thrown horizontally, so there is no initial vertical velocity)
- g is the acceleration due to gravity (-9.8 m/s^2)
- t is the time we want to solve for
Since the baseball is thrown horizontally, the initial vertical velocity v0y is zero. The equation simplifies to:
y = y0 - (1/2)gt^2
Plugging in the values we have:
40 = 0 - (1/2)(-9.8)t^2
Simplifying further:
40 = (4.9)t^2
Now, we can solve for t by isolating it:
t^2 = 40 / 4.9
t^2 = 8.1632653
Taking the square root of both sides to solve for t:
t = sqrt(8.1632653)
t ≈ 2.86
So, it will take approximately 2.86 seconds for the baseball to hit the ground.