A ball is thrown straight upward and rises to a maximum height of 25 m above its launch point. At what height

Question

has the ball decreased to one-half of its initial value?

Answer 1

Let's assume no air resistance and approximate g as -10 m/s/s. Since the acceleration is constant the speed will be half its initial value at one-half the time to the top. But since the distance moved by an accelerating object depends on the square of the time, the ball will be past half way to the top at half the time. The height should be 25 m /(sqr rt of 2). That would be about 18 m. The equation for height is h = Vot -1/2 g t^2.

Answer 2

eeshk. Supposing there's no auxilary wind movement and ignoring the minute air pressure difference over the 25m in height, I'd have to say exactly 12.5m up.

However, I realise while typing, that wind resistance decreases in proportion to the decrease of speed, so in that case we'd be talking more of a half life of the speed.

Answer 3

so this is basically a 1-dimensional kinematics problem

therefore the motion of ball going up and going down (neglecting air resistance) will form a parabola and will be symmetrical on path up and down

time it takes to fall from max height:

d=0.5at^2

-25=0.5(-9.8)t^2

t= 2.258769757 s

calculate its impact speed, remember since motion is symmetrical that means its impact speed will be the same as its initial speed (no loss of energy)

a=v/t

9.8=t/2.258769757

v= 22.13594362 m/s

and half of that value would be 11.06797181 m/s

t= 11.06797181/9.8= 1.129384879 seconds to reach that speed

so it would travel

d=0.5at^2

d=0.5(-9.8)(1.129384879^2)

d= 6.25 metres from max height

from the launch point it will be= 25-6.25= 18.75 metres

so it will decrease to half its initial speed value at 18.75 metres

hope this helps