G) www.webassign net 40 m/s to -40 m/s in steps of -10 m/s every second: hence,
ID: 1621182 • Letter: G
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G) www.webassign net 40 m/s to -40 m/s in steps of -10 m/s every second: hence, approximately 8 seconds elapse during the motion. The velocity vector constantly changes direction, but the horizontal velocity never changes because the acceleration in the horizontal direction is zero. Therefore, the displacement of the ball in the x-direction is given by the equation, Ar a vor1 (20 m/s)(8 s) 160 m. LEARN MORE REMARKS This example emphasizes the independence of the x- and y components in projectile motion problems QUESTION How does the magnitude of the velocity vector at impact compare with the magnitude of the initial velocity vector? It is greater initially. O 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. O 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. PRACTICE IT 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 30 m/s and 1sm/s, respectively. use a motion diagram to estimate the ball's total time of flight and the distance it traverses before hitting the ground (use 10 m/s the acceleration of gravity.) Distance EXERCISE HINTS: GETTING STARTED I IM STUCK! A ball is thrown so that its initial vertical and horizontal components of velocity are 30 m/s and 15 m/s, respectively. Estimate the maximum height the ball reaches. (use 10 m/s2 as the acceleration of Need Help? to 7, -3 points seroPg 3 AE 008 EXAMPLE 3.8 l at's Ouite an ArmExplanation / Answer
PRACTICE IT:
vy = 30 m/s
vx = 15 m/s
in vertical, as it comes back to its initial vertical position. (ground)
y = 0
y = vy t + ay t^2 /2
0 = 30 t - 9.8 t^2 / 2
t = 6.12 sec .........Ans(Time of flight )
In horizontal, there is no acceleration.
hence Dx = vx t
Dx = 15 x 6.12 = 91.8 ..Ans
EXERCISE :
at maximum height, vfy= 0
Applying v^2 - u^2 = 2 a d
0^2 - 30^2 = 2 (-9.8) Hmax
Hmax = 45.9 m ........Ans
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