Question

rock is thrown vertically upward from the ground with an initial speed 15m/s. a. how high does it go b. how much time is required for the rock to reach its maximum height? c. what is the rock’s height at t=2.00s?​

Answer 1

initial velocity(u)=15m/s

final velocity(v)=0

acceleration=-10m/s^2

so,a.

2as=(v)^2-(u)^2

=>2×(-10)×s=0-(15)^2

=>-20s=0-225=-225

=>s=-225/-20

=>s=45/4=11.25meters

therefore,it goes 11.25 meters high.

so,b.

v=u+at^2

=>0=15+(-10)×t^2

=>0=15-10×t^2

=>0=5×t^2

=>t^2=0-5=-5

=>t=√-5 secs

therefore,it takes √5 secs to cover 11.25 meters.

sorry I'm not clear bout' no.c....

Answer 2

Answer:

Explanation:

initial velocity(u)=15m/s

final velocity(v)=0

acceleration=-10m/s^2

so,a.

2as=(v)^2-(u)^2

=>2×(-10)×s=0-(15)^2

=>-20s=0-225=-225

=>s=-225/-20

=>s=45/4=11.25meters

therefore,it goes 11.25 meters high.

so,b.

v=u+at^2

=>0=15+(-10)×t^2

=>0=15-10×t^2

=>0=5×t^2

=>t^2=0-5=-5

=>t=√-5 secs

Repeat these for c one also