a ball is thrown vertically upward with a velocity of 19.6 m/s. find maximum height reached by the ball and time taken to reach maximum height? what is the velocity of the ball at t=3s?
Answers
Answer:
Final velocity (after 3 secs) = initial velocity(19.6 m/s) + acceleration of gravity (-9.8 m/s^2) multiplied by time (3 secs).
V=u+at=19.6 + (-9.8) (3)= 19.6- 29.4= -9.8 meters/second
Initial velocity (when thrown) was 19.6 m/s upward, the positive direction.
Zero velocity (at max height) was attained at 19.6 meters height and 2 seconds time.
Final velocity (after 3 seconds) was 9.8 m/s downward, the negative direction.
Maximum height attained is found by turning the ball’s initial kinetic energy into potential energy:
mgh=1/2mv^2 (mass cancels out)
9.8 h=1/2 (19.6)^2= 192
h=19.6 meters= max height
Height after 3 seconds is height after 1 second of downward travel from max height.
s= ut + 1/2 at^2= 0 + 1/2 (9.8) 1^2= 4.9 meters below max height.
h-s= 19.6–4.9= 14.7 meters above point where thrown.