A .45-caliber bullet, shot straight up from the surface of the moon would reach a height of s (1) = 8321 - 2.61
feet after t-seconds. On Earth, in the absence of air, its height would be s(r) = 832t - 167- feet after t
seconds. How long would it take for the bullet to hit the surface on both bodies and at what speed does it hit?
To find out when the bullet hits the surface on both the moon and Earth, we need to set the height equal to zero and solve for t.
On the moon:
8321 - 2.61t = 0
2.61t = 8321
t = 8321/2.61
t ≈ 3188.89 seconds
On Earth:
832t - 167 = 0
832t = 167
t = 167/832
t ≈ 0.20 seconds
Now let's calculate the speed at which the bullet hits the surface on both bodies.
On the moon:
The velocity of the bullet when it reaches the highest point (h=0) is given by the derivative of the height function. Taking the derivative of s(1) = 8321 - 2.61t, we get:
v(1) = -2.61 ft/s
The negative sign indicates that the bullet is moving downwards.
On Earth:
The velocity of the bullet when it reaches the highest point (h=0) is also given by the derivative of the height function. Taking the derivative of s(r) = 832t - 167, we get:
v(r) = 832 ft/s
Therefore, the bullet hits the surface on the moon with a speed of -2.61 ft/s (moving downwards), and it hits the surface on Earth with a speed of 832 ft/s.