A ball is thrown vertically upward with a velocity of 98 metre per second how high does the ball rise
Answers
Answer:xf=xi + vi(t) + (at^2)/2
Make the vertical axis the x-axis. Could have used y, but I don't want to retype the formula.
xf= 0 + 98(t) -4.9t^2
vf= vi + at ….the ball is projected up. Once the ball leaves the cannon, only gravity and air resistance act upon it. We’re neglecting air resistance, so we’re considering only gravity. As the ball rises it continues to decelerate until it reaches its maximum height, at which point it momentarily stops, and its velocity is zero. So, vf = 0. We couldn't make this assumption if the ball was dropped, since we don't know its velocity when it hits the ground.
0 = 98 - 9.81(t)
t= 10 s
xf = 980 -490 = 490m This is the maximum height.
Explanation:
Answer:
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Explanation:
As we know, v² = u²-- 2gh
(0)² = u² – 2gh
u² = 2gh
(98)² = 2 x 9.8 x h
98 x 98 ÷ 2 x 9.8 = h
h = 98 x 98 x10 ÷ 2 x 98
h = 98 x 5
h = 490 m
As we know, v = u + gt
(0)² = 98 + (– 9.80) x t
98 = 9.8 t
t = 98 ÷ 9.8
t = 10 sec.
Total time = time to go up + time to go down
= 10 + 10 = 20 sec