Posted by **Letisha** on Wednesday, November 25, 2009 at 6:10pm.

A man can throw a ball a maximum horizontal

distance of 97.8 m.

The acceleration of gravity is 9.8 m/s2 .

How far can he throw the same ball verti-

cally upward with the same initial speed?

Answer in units of m.

- physics -
**MathMate**, Wednesday, November 25, 2009 at 9:06pm
The maximum horizontal distance is obtained by throwing the ball at 45° with the horizontal.

Let v0=initial speed,

resolve v0 into horizontal and vertical components,

vh=v0*cos(45)

vv=v0*sin(45)

The time, t, the ball was in the air can be calculated by the horizontal distance divided by the horizontal component of velocity, vh.

t=97.8/vh

The vertical distance can be calculated using

S=vv*t+(1/2)gt²

where g=-9.8 m/s² (downwards)

Substitute vv and t into the above equation gives

S=(v0 sin45)*(97.8/(v0 cos45) + (1/2)g (97.8/(v0 cos45))²

From which v0 can be calculated as approximately 31 m/s.

Now apply the same velocity in the vertical direction,

initial kinetic energy

KE = (1/2)mv0²

At the top, h metres above ground, kinetic energy is zero, but potential energy is

PE = mgh

Equate KE=PE,

mgh=(1/2)mv0²

h=(1/2)mv0²/(mg)

=(1/2)v0²/g

= 49 m, approx.

- physics -
**Damon**, Wednesday, November 25, 2009 at 9:21pm
First you need to know what angle from horizontal gives maximum range. With no air friction it is 45 degrees but you need to show that.

V = throwing speed

u = V cos A

Vo = V sin A

R = range = u T = V T cos A = 97.8

speed up = Vo - (9.8) t

= 0 at top where t = T/2 because half the time is rising and half falling

so

0 = Vo - 4.9 T

V sin A = 4.9 T

so T = .204 V sin A

range = V T cos A

range = .204 V^2 sin A cos A

so for what angle A is sin A cos A a maximum?

y = sin x cos x

dy/dx = 0 at max = -sin^2 x + cos^2 x

that is where the sin is equal to the cosine, 45 degrees, where sin = cos = .707 and sin*cos = .707*.707 = .5

so now we do the problem

range = .204 V^2 (.5) = 97.8

V^2 = 959

31 m/s approx

NOW throw vertical up at 31 m/s

31 = 9.8 t at top

t = 3.16 seconds traveling up

h = V t - 4.9 t^2

= 31*3.16 - 4.9(3.16^2)

= 97-49 = 48 meters approximately

- physics -
**Damon**, Wednesday, November 25, 2009 at 9:23pm
MathMate's energy method is far cooler than my more basic method.

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