A stone is thrown straight down from the edge of a roof, 725 feet above the ground, at a speed of 7 feet per second. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 3 seconds later? At what time does the stone hit the ground? What is the velocity of the stone when it hits the ground?
+33654
1. Use s = ut + (1/2)at2, where s = distance travelled, u = initial velocity, a = acceleration, t = time.
s = 7*3 +(1/2)*32*9 = 165 ft., so the stone is 725 - 165 = 560 ft off the ground.
2. The stone hits the ground when s = 725, so: 725 = 7*t +(1/2)*32*t2
Solving this quadratic we get
16×t2+7×t−725=0⇒{t=−(3×√5161+7)32t=(3×√5161−7)32}⇒{t=−6.9537593958731787t=6.5162593958731787}
The only positive solution is t = 6.516 seconds
3. Use v2 = u2 + 2as where v is final velocity.
v2 = 49 + 2*32*725
v=√49+2×32×725⇒v=215.5203006679417195
v = 215.52 ft/sec
.
+33654
1. Use s = ut + (1/2)at2, where s = distance travelled, u = initial velocity, a = acceleration, t = time.
s = 7*3 +(1/2)*32*9 = 165 ft., so the stone is 725 - 165 = 560 ft off the ground.
2. The stone hits the ground when s = 725, so: 725 = 7*t +(1/2)*32*t2
Solving this quadratic we get
16×t2+7×t−725=0⇒{t=−(3×√5161+7)32t=(3×√5161−7)32}⇒{t=−6.9537593958731787t=6.5162593958731787}
The only positive solution is t = 6.516 seconds
3. Use v2 = u2 + 2as where v is final velocity.
v2 = 49 + 2*32*725
v=√49+2×32×725⇒v=215.5203006679417195
v = 215.52 ft/sec
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