Posted by Letisha on Wednesday, November 25, 2009 at 6:10pm.
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.
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
MathMate's energy method is far cooler than my more basic method.