Explanation:
1. CS ≅ HR, ∠CHS ≅ ∠HCR, ∠CSH ≅ ∠HRC — given
2. ∆CRH ~ ∆HSC — AA similarity theorem
3. ∠SCH ≅ ∠RHC — corresponding angles of similar triangles are congruent
4. CH ≅ HC — reflexive property of congruence
5. ∆CRH ≅ ∆HSC — SAS congruence theorem
6. CR ≅ HS — CPCTC
Answer:
(a) 2.79 seconds after its release the bag will strike the ground.
(b) At a velocity of 73.28 ft/second it will hit the ground.
Step-by-step explanation:
We are given that a balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 80 feet above the ground.
Assume the acceleration of the object is a(t) = −32 feet per second.
(a) For finding the time it will take the bag to strike the ground after its release, we will use the following formula;
Here, s = distance of the balloon above the ground = - 80 feet
u = intital velocity = 16 feet per second
a = acceleration of the object = -32 feet per second
t = required time
So,
Now, we will use the quadratic D formula for finding the value of t, i.e;
Here, a = 1, b = -1, and c = -5
Also, D = =
= 21
So,
We will neglect the negative value of t as time can't be negative, so;
= 2.79 ≈ 3 seconds.
Hence, after 3 seconds of its release, the bag will strike the ground.
(b) For finding the velocity at which it hit the ground, we will use the formula;
Here, v = final velocity
So,
v = 16 - 89.28 = -73.28 feet per second.
Hence, the bag will hit the ground at a velocity of -73.28 ft/second.