posted by Letisha on .
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.
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,
The time, t, the ball was in the air can be calculated by the horizontal distance divided by the horizontal component of velocity, vh.
The vertical distance can be calculated using
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
= 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
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.