a joggler tosses 3 balls alternately vertically upward. each ball has an initial velocity of 5m/s. how high does each ball rise? how long does each ball remain in the air?
please with solution...thanks...i cant solve this
2007-01-22 21:52:20 · 4 answers · asked by antoinekingx 1 in Science & Mathematics ➔ Physics
You need to know the shape, size and material of construction of the balls to determine the losses due to air resistance.
2007-01-22 22:19:50 · answer #1 · answered by tgypoi 5 · 0⤊ 0⤋
You use the equation v = u + at to calculate how long the ball takes to reach its highest point, at which its velocity will be zero. You then double the time to see how long the ball was in the air.
You then use the equation s = ut + ½at² to calculate the distance that the ball travels in half the period of time.
In each case, "a" is acceleration due to gravity, 9.8 m/s².
The balls are in the air for just over a second and rise to a height of just under 4m.
2007-01-22 22:19:19 · answer #2 · answered by Ben C 2 · 0⤊ 0⤋
Use the equation v^2 = u^2 + 2* g * h
When the ball attains its max height, v = 0; furthermore, g = - 9.81, u = 5 m/s; substituting gives h = 1.27m
The time taken to reach this height is got from eqn v = u + g* t, giving t = 5/9.81 . Total time taken = 2* 5/9.81 = 1.02 sec
2007-01-22 22:35:52 · answer #3 · answered by Paleologus 3 · 0⤊ 0⤋
a =-10
u=5
v=0
a=(v-u)/t
so t=(v-u)/a
t=(0-5)/-10
=0.5s
s=ut
=(5)(0.5)
=2.5m
2007-01-22 22:23:43 · answer #4 · answered by ask_n_answer 1 · 0⤊ 0⤋