This problem involves the use of a kinematic equation since it involves the motion of an object. The equation for the height of the object is given as:
s = -<span>16*t^2+v_o*t
Also, the initial velocity, v_o, was also said to be equal to 128 ft/s.
For the first question, </span><span>the time(s) that the projectile will reach a height of 240 ft when v_o is 128 feet per second, can be solved using the given equation and the quadratic formula. The resulting quadratic equation is then </span>-16*t^2 + 128*t -240 =0, where a =16, b =128, and c =-240. The quadratic formula is equal to [-b <span>± sqrt(b^2 -4ac)]/2a. This gives two answers t = 3 seconds and t = 5 seconds. This might be because the projectile has a parabolic path, thus, it reaches the height of 240 ft, before and after it reaches a peak.
For the second question, the time it takes for the projectile to reach the ground is obtained by setting the distance, s, equal to zero. In this case, t = 8 seconds.</span>
3x + 3y = 27
3x - 3y = -11
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6x = 16
x = 16/6 which reduces to 8/3
3x + 3y = 27
3(8/3) + 3y = 27
8 + 3y = 27
3y = 27 - 8
3y = 19
y = 19/3
solution is (8/3, 19/3)
Answer: Neither, they are the same.
Answer:
47*
Step-by-step explanation:
Using the vertical angle rule, I can get CBD as 47*
Answer:
112
Step-by-step explanation:
7+9+11+13+15+17+19+21