Final answer:
We solve the problem with the quadratic formula, taking into account the downward initial velocity, the acceleration due to gravity, and the height of the tower. With the negative root of the square root involved in the calculations, we determine the time at which the velocity reaches -5.42 m/s, which in turn allows us to find the time the penny hits the ground.
Step-by-step explanation:
To determine when the penny hits the ground, we can use the equations of motion for an object under constant acceleration, due to gravity, in a straight line. The relevant equation for this scenario, where we know the initial velocity (vi), the acceleration due to gravity (g), and the displacement (height of the tower), is:
s = vit + (1/2)gt^2
where:
- s is the displacement (height from which the penny is thrown, which is 90 m in this case)
- vi is the initial velocity (in this case, -5 m/s, as the penny is thrown downward)
- g is the acceleration due to gravity (approximately -9.8 m/s^2, with downward direction being negative)
- t is the time we are solving for
To solve for time (t), we rearrange the equation and plug in the known values:
0 = 90 - 5t - (1/2)(9.8)t^2
This is a quadratic equation in the form of at^2 + bt + c = 0, which can be solved for time t using the quadratic formula. However, since we are given some intermediate steps, we first find the square root of 29.4 and, choosing the negative root, -5.42 m/s as it is the final velocity directed downwards when hitting the ground. We then have an intermediate step where Vy = -5.42 m/s.
The final step is to use this velocity in our equation to solve for t, giving us the exact time when the penny reaches the ground.