Answer:
1. t= 0.98
2. v(t)= -9.8t
3. a(t)= -9.8
4. v(0.98)= -9.604 m/s
5. |v(0.98)|= 9.604
6. a(0.98)= -9.8 m/s²
Explanation:
To find out when the ball goes splat on the ground, and then use some fancy math to figure out how fast and furious it was falling, follow these steps:
1. The ball goes splat when its height is zero, so we need to solve h(t) = 0. Using the quadratic formula, which you probably learned in high school and then forgot, we get:
t = (-b ± √(b² - 4ac)) / (2a)
where a = -4.9, b = 0 and c = 94. Plugging in these numbers, we get:
t = (-0 ± √(0² - 4(-4.9)(94))) / (2(-4.9))
t = (0 ± √(1846.8)) / (-9.8)
t = (0 ± 42.96) / (-9.8)
t = -4.38 or 0.98
We ignore the negative value because it means the ball was already on the ground before we threw it, which is silly. So the time when the ball goes splat is t = 0.98 seconds.
2. The speedometer of the ball at any time t is given by the slope of h(t) with respect to t:
v(t) = h’(t) = -9.8t
v(t) = -9.8t
This is a straight line that shows that the speedometer of the ball is going down at a steady rate of -9.8 m/s².
3. The gas pedal of the ball at any time t is given by the slope of v(t) with respect to t:
a(t) = v’(t) = -9.8
a(t) = -9.8
This is a flat line that shows that the gas pedal of the ball is always -9.8 m/s², which is equal to the force that makes things fall on Earth.
4. The speedometer of the ball at the moment it goes splat is given by plugging in t = 0.98 into v(t):
v(0.98) = -9.8(0.98)
v(0.98) = -9.604
Therefore, the speedometer of the ball at the moment it goes splat is -9.604 m/s.
5. The actual speed of the ball at the moment it goes splat is given by taking off the minus sign from its speedometer:
|v(0.98)| = -9.604
|v(0.98)| = 9.604
Therefore, the actual speed of the ball at the moment it goes splat is 9.604 m/s.
6. The gas pedal of the ball as it goes splat is given by plugging in t = 0.98 into a(t):
a(0.98) = -9.8
Therefore, the gas pedal of the ball as it goes splat is -9.8 m/s².