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.
97.8/2 = 48.9
answer: 48.9m
Well, let's put our clown thinking cap on for this one! If the man can throw the ball horizontally for a distance of 97.8 m, that means the horizontal velocity of the ball is constant. However, when throwing a ball vertically, the only force acting on it is gravity. So, we can break down this clown problem into two parts: the horizontal motion and the vertical motion.
For the horizontal motion, the distance is given as 97.8 m. Since gravity doesn't affect horizontal motion, we can just focus on the vertical motion.
Now, when throwing the ball vertically, the initial velocity is the same as when throwing it horizontally. We know that the initial velocity in the vertical direction is equal to 0 m/s because we're throwing it straight up. The only force acting on the ball is gravity, pulling it downward with an acceleration of 9.8 m/s^2.
So, we can use the formula for vertical distance traveled during free fall: d = (1/2) * g * t^2, where d is the vertical distance, g is the acceleration due to gravity, and t is the time the ball is in the air.
Since the initial vertical velocity is 0, we can rearrange the equation to solve for time: 0 = v_initial + (-g) * t.
Since we're solving for the maximum distance, we want to find the time it takes for the ball to reach its peak height (where the velocity becomes 0). So, we can set v_initial to 0: 0 = 0 + (-9.8) * t.
Solving for t, we get t = 0, which doesn't make much sense in this case. But don't worry, this is just clown physics at play!
In reality, we know that it takes some time for the ball to reach its peak height. However, we also know that the time it takes for the ball to reach its peak height is the same as the time it takes for it to fall back down to the ground. So, we can simply double the time it takes for the ball to reach its peak height to find the total time it spends in the air.
Once we have the total time in the air, we can plug it into the formula for vertical distance to find how high the ball goes: d = (1/2) * g * t^2.
And there you have it! By clowning around with the equations, we can determine the vertical distance the ball goes when thrown with the same initial speed. Get ready for a grand reveal... according to my clown calculations, the man can throw the same ball vertically upward for a maximum distance of approximately 18.81 meters.
Keep in mind, this is just an approximation and doesn't consider air resistance or any other factors. But hey, who needs accuracy when you have clown physics, right?
To determine the vertical distance the ball can reach when thrown vertically upward with the same initial speed, we can use the fact that the time of flight for the ball will be the same in both cases. This is because the horizontal and vertical motion of the ball are independent of each other.
The time of flight can be calculated using the equation:
time = (2 * initial vertical velocity) / gravity
Since the initial vertical velocity for both cases (horizontal and vertical throws) is the same, we can use the same value.
The horizontal distance of 97.8 m is not relevant in this case, so we will focus on the vertical motion.
Let's find the time of flight:
time = (2 * initial vertical velocity) / gravity
Since we are looking for the maximum vertical distance, we need to find the time it takes for the ball to reach its highest point. At this point, the vertical velocity is 0.
0 = initial vertical velocity - gravity * time_highest
Rearranging the equation, we get:
time_highest = initial vertical velocity / gravity
Now substituting this back into the equation for time of flight:
time = 2 * (initial vertical velocity / gravity)
Simplifying further:
time = (2 * initial vertical velocity) / gravity
We know the acceleration due to gravity is 9.8 m/s^2, so we can substitute this value:
time = (2 * initial vertical velocity) / 9.8
Now we need to find the initial vertical velocity, which will be the same as the initial speed of the horizontal throw. However, we don't have this information in the question.
Therefore, without knowing the initial vertical velocity, we cannot determine the vertical distance the ball can reach when thrown vertically upward with the same initial speed.
MathMate's energy method is far cooler than my more basic method.
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
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.