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.
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?
1 answer