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anyanavicka [17]
3 years ago
9

a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik

es the ground. The initial velocity of the ball is 82 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to _____
Physics
1 answer:
grin007 [14]3 years ago
7 0

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:  

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)  

Where:  

y=0 is the height of the ball when it hits the ground  

y_{o}=70 m is the initial height of the ball

V_{o}=82m/s is the initial velocity of the ball  

t is the time when the ball strikes the ground

g=9.8m/s^{2} is the acceleration due to gravity  

Having this clear, let's find t from (1):  

0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2} (2)  

Rewritting (2):

-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0 (3)  

This is a quadratic equation (also called equation of the second degree) of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}  (4)

Where:

a=-\frac{1}{2}(9.8m/s^{2}

b=82m/s

c=70m

Substituting the known values:

t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}  (5)

Solving (5) we find the positive result is:

t=17.68 s

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Answer:

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Explanation:

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Using the formula for the acceleration we find

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