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Now let’s consider a projectile problem in which the initial velocity is specifi

ID: 1362614 • Letter: N

Question

Now let’s consider a projectile problem in which the initial velocity is specified in terms of a magnitude and an angle. Suppose a home-run baseball is hit with an initial speed v0=37.0m/s at an initial angle ?0=53.1?. (a) Find the ball’s position, and the magnitude and direction of its velocity, when t=2.00s. (b) Find the time the ball reaches the highest point of its flight, and find its height h at that point. (c) Find the horizontal range R (the horizontal distance from the starting point to the point where the ball hits the ground).

Part A - Practice Problem:

If the ball could continue to travel below its original level (through an appropriately shaped hole in the ground), then negative values of y corresponding to times greater than 6.04 s would be possible. Compute the x-component of the ball's position 7.60 s after the start of its flight.

Part B - Practice Problem:

Compute the y-component of the ball's position 7.60 safter the start of its flight.

Part C - Practice Problem:

Compute the x-component of the ball's velocity 7.60 safter the start of its flight.

Part D - Practice Problem:

Compute the y-component of the ball's velocity 7.60 safter the start of its flight.

Figure l # 1 of 1 y (m) v =37.0m/s h- is = R = ? 12-

Explanation / Answer

vo =37 m/s, g =9.8 m/s^2

vyx =37 cos(53.1) =22.2 m/s

voy = 37 sin(53.1) = 29.6 m/s

PartA : t = 7.6 s

x =(vox)t = 22.2*7.6

x =168.72 m

PartB : for y it is moving in negative direction. so t =7.6 -6.04 =1.56s

y =(voy)t+(1/2)at2

y= (29.6)(1.56)-(1/2)(9.8)(1.56)2

y= 34.276 along down ward

Part C:

vx =vox = 22.2 m/s

Part B:

vy = voy+at

vy = 29.6-(9.8*1.56)

vy = 14.3 m/s along down ward direction

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