A rock is dropped from a 100 foot tower. The height of the rock as a function of time can be modeled by the equation: h(t) = -16t 2 + 100. How long does it take for the rock to reach the ground?
It takes 2.5 seconds to reach the ground. Problem 2 : A rock is dropped on the surface of Mars from a height of 100 feet. The height of a falling rock as a function of time since it was dropped on Mars can be modeled by the equation: h(t) = -6.5t 2 + 100. How long does it take for the rock to hit the surface of Mars? Solution: h(t) = -6.5t 2 + 100 h(t) = 0 0 = -6.5t 2 + 100 6.5t 2 = 100
So, it takes 3. Problem 3 : A ball is thrown from ground level upward at an initial velocity of 60 ft/sec. What is the ball's maximum altitude? The equation for "projectile motion" is h(t) = -16t 2 + 60t. Solution: h(t) = -16t 2 + 60t a = -16, b = 60
h(1.875) = -16(1.875) 2 + 60(1.875) h(1.875) = -56.25 + 112.5 h(1.875) = 56.25 feet So, maximum altitude is 56.25 feet. Problem 4 : A ball is thrown upward from the surface of Mars with an initial velocity of 60 ft/sec. What is the ball's maximum height above the surface before it starts falling back to the surface? The equation for "projectile motion" on Mars is: h(t) = -6.5t 2 + 60t Solution: h(t) = -6.5t 2 + 60t
h(4.6154) = -6.5(4.6154) 2 + 60(4.6154) = -138.46 + 276.92 h(4.6154) = 138.46 feet So, the maximum height is 138.46 feet. Problem 5 : A rock is thrown upward from ground level with an initial velocity of 50 feet/sec. When will the rock hit the ground? Projectile motion can be modeled by the equation: h(t) = -16t 2 + 50t. Solution: h(t) = -16t 2 + 50t h = 0 0 = -16t 2 + 50t 0 = -2t(8t - 25)
| -2t = 0 t = 0 | 8t - 25 = 0 8t = 25 t = 25/8 t = 3.125 sec |
So, in 3.125 sec the rock hits the ground. Problem 6 : A rock thrown upward from the surface of Mars with an initial velocity of 50 feet per second. The height of a rock can be modeled by the: h(t) = -6.5t 2 + 50t. How long does it take the rock to fall back to the surface of Mars? Solution: h(t) = -6.5t 2 + 50t h = 0 0 = -6.5t 2 + 50t 0 = t(-6.5t + 50)
| t = 0 | -6.5t + 50 = 0 6.5t = 50 t = 50/6.5 t = 7.612 sec |
So, it will take 7.612 sec to fall back to the surface of Mars. Problem 7 : A rock is thrown upward from the top of a 25 foot tower with an initial upward velocity of 100 ft/sec. The height of a rock above the ground as a function of time can be modeled by the equation: h(t) = -16t 2 + 100t + 25. How long does it take for the rock to: a) reach its maximum height? b) fall back to the ground? Solution:
a) h(t) = -16t 2 + 100t + 25