A body is thrown with a
relocity of 9.8mls
vertically up. How much time it takes to strike the ground
Answers
Answer:
Always watch your sign convention when using Newton’s equations of motion. I always assume positive = up and negative = down, so the acceleration of gravity is *always* negative:
ay=−9.81ms2
Begin with the following equation of motion:
S=(vo)t+12at2
Since we have motion in both the x-direction and the y-direction, write Newton’s equation in the y-direction since we are interested in how long the object is in the air. Use subscripts to help:
Sy=(vo)yt+12ayt2 ————————— eqn 1
If the object landed below the starting point, then Sy would be a negative number. In your problem, the object lands back at the starting point, so Sy=0
Also, since the initial velocity is upward, (vo)y is positive.
Equation 1 becomes:
0=9.8(sin30)t+12(−9.81ms2)t2
divide each term by “t”
0=9.8(sin30)+12(−9.81ms2)t
t = 1 sec.
Explanation:
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