A ball is thrown up with a speed of 0.5 m/s.take the value of accleration due to gravity 9.8m/s .how high will it gobefore it begain to fall?how long it will take to reach that height.
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Here, u=0.5 m/s, v=0, s=?, t=?
Let us take, a=−10 m/s2 in vertical upward motion of the ball upto highest point.
Using the third equation of motion, we get
v2=u2+2as or, s=v2−u22a =0−0.522×(−10)=0.0125 m
Now, using the first equation of motion, we get
v=u+at0=0.5+(−10) tor, t=0.05 s
Let us take, a=−10 m/s2 in vertical upward motion of the ball upto highest point.
Using the third equation of motion, we get
v2=u2+2as or, s=v2−u22a =0−0.522×(−10)=0.0125 m
Now, using the first equation of motion, we get
v=u+at0=0.5+(−10) tor, t=0.05 s
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