Question

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point a distance 48.0 m below its starting point at a time 5.00s after it leaves the thrower's hand. Air resistance may be ignored.


part A:

What is the initial speed of the egg?

v= m/s


part B:

How high does it rise above its starting point?

h= m


part C:

What is the magnitude of its velocity at the highest point?

v= m/s


part D:

What is the magnitude of its acceleration at the highest point?

a= m/s2


part E:

What is the direction of its acceleration at the highest point?

up or down

Answer 1

Use the kinem atic eqauti on of motion to solve for initial speed. -48- o (5.00 s)+-9.8 m/s2) (5.00 s) -48 m + 122. 5 m-149m/s] oy5.00 s Use the kinem atic eqauti on of moti on to the maximum height. -0,-25 0-(14.9m/s) -98m/s) 11.32 m The velocity of the egg is zero at the maximum height The accelerati on will always be directed downward, that because it is the accelerati on due to gravity, which is uqual to 9.8 m/s the accelerati on will stay the same at any point of a free-fall motion.