Question

question how does the magnitude of the velocity vector at impact compare with the magnitude of the initial velocity vector?
it is greater at impact.
they are the same since the magnitude of the vertical component of velocity is the same at each height on the way up and on the way down.
they are the same since the ball during its flight has an upward acceleration as its height increases and a downward acceleration on its way down.
it is greater initially.
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use the worked example above to help you solve this problem. a ball is thrown so that its initial vertical and horizontal components of velocity are 50 m/s and 25 m/s, respectively. use a motion diagram to estimate the balls total time of flight and the distance it traverses before hitting the ground. (use 10 m/s² as the acceleration of gravity.)
time s
distance m
exercise
hints: getting started | im stuck!
a ball is thrown so that its initial vertical and horizontal components of velocity are 50 m/s and 25 m/s, respectively. estimate the maximum height the ball reaches. (use 10 m/s² as the acceleration of gravity.) m
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Explanation:

Step1: Analyze vertical - motion time

The vertical - motion of the ball is a free - fall motion. The time it takes for the ball to reach its maximum height can be found using the formula $v = v_0+at$. At the maximum height, the vertical velocity $v = 0$. Given $v_0 = 50\ m/s$ and $a=- 10\ m/s^2$, we have $0 = 50-10t_1$, so $t_1=\frac{50}{10}=5\ s$. The total time of flight $t$ is twice the time to reach the maximum height, so $t = 2t_1$.
$t = 2\times5=10\ s$

Step2: Calculate horizontal distance

The horizontal motion of the ball is a uniform - motion with constant velocity. The horizontal velocity $v_x = 25\ m/s$. The horizontal distance $x$ is given by the formula $x = v_x t$. Substituting $v_x = 25\ m/s$ and $t = 10\ s$ into the formula, we get $x=25\times10 = 250\ m$.

Step3: Calculate maximum height

For the vertical motion to find the maximum height $h$, we use the formula $v^2 - v_0^2=2ah$. At the maximum height $v = 0$, $v_0 = 50\ m/s$ and $a=-10\ m/s^2$. Then $0 - 50^2=2\times(- 10)\times h$. Solving for $h$ gives $h=\frac{50^2}{2\times10}=\frac{2500}{20}=125\ m$.

For the multiple - choice question:
The magnitude of the velocity vector at impact is the same as the magnitude of the initial velocity vector since the magnitude of the vertical component of velocity is the same at each height on the way up and on the way down (neglecting air - resistance). The horizontal component of velocity remains constant throughout the motion.

Answer:

They are the same since the magnitude of the vertical component of velocity is the same at each height on the way up and on the way down.
Time: 10
Distance: 250
Maximum height: 125