A ball is thrown from an initial height of
1 answer:
Answer:
Step-by-step explanation:
If we are looking for the times that the ball was 11 meters off the ground, we sub in 11 for the height on the left and solve for t:
and
and factor this however it is you are factoring in class to solve for t to get
t = .35 seconds and t = 5.6 seconds
Because the ball reaches this point in its way up and then passes it again on its way down, the ball will have 2 times at this height.
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Answer:
Given: BD is an altitude of △ABC .
Prove: sinA/a=sinC/c
Triangle ABC with an altitude BD where D is on side AC. Side AC is also labeled as small b. Side AB is also labeled as small c. Side BC is also labeled as small a. Altitude BD is labeled as small h.
Statement Reason
BD is an altitude of △ABC .
Given △ABD and △CBD are right triangles. (Definition of right triangle)
sinA=h/c and sinC=h/a
Cross multiplying, we have
csinA=h and asinC=h
(If a=b and a=c, then b=c)
csinA=asinC
csinA/ac=asinC/ac (Division Property of Equality)
sinA/a=sinC/c
This rule is known as the Sine Rule.
