Assuming that accelerations is constant,
Given: a=15 ms^-2, s=14.1 m
Applying kinematics equation,
v^2 = u^2 + 2as
Find for u since v is 0,
u= sqrt(2 x -15 x -14.1)
= 20.78
Solve for t,
s= ut + 0.5at^2
14.1= 20.78t + 0.5(15)t^2
//Solve quadratic eqn for t
t= 0.564(Ans), -3.33(not possible)
Okay so the acceleration g on this planet X is 15 m/s^2
I have solved for all answers but this one.
How long would it take the object to reach a height of 14.1 ? For an object thrown upward there are two times it would be at any hieght. Give the going up answer, a comma, and give the going down answer.
How do we go about this?
2 answers
Solving for t,
14.1=20.78t-.5(15)t^2
subtract using a -15m/s^2 (downward acceleration due to gravity)
14.1=20.78t-.5(15)t^2
subtract using a -15m/s^2 (downward acceleration due to gravity)
1 answer