A baseball is thrown horizontally off a cliff with a speed of 20 m/s. What is the horizontal distance, to the nearest tenth of a meter, from the face of the cliff after 4.9 seconds? 8 In the previous problem, to the nearest tenth of a meter, how far has it fallen in that time?

a) 98 m

b) 117.8 m

Explanation:

Hi!

Since there is no force acting on the horizontal direction, the horizontal displacement x as a function of time t is:

x (t) = 20 m/s * t

Therefore, the horizontal distance after 4.9 seconds is:

x (4.9 s) = (20 m/s) * (4.9 s) = 98 m

On the other hand, on the vertical direction, there is no initial velocity, but a constant acceleration g is acting on this direction, therefore, the vertical displacement is given by:

y (t) = - (1/2) * g*t^2

Therefore, the vertical distance after 4.9 seconds is:

y (4.9 s) = - (1/2) * (9.81 m/s^2) * (4.9 s) ^2 = - 117.769 m = - 117.8 m