You are outside in a big, level field, in the midst of which stands a narrow wal

Question

You are outside in a big, level field, in the midst of which stands a narrow wall, 33.2 m high. You are given a device that can launch a projectile, always with the same speed of 58.0 m/s.

a) first, you launch a porjectile from the top of the wall, at an angle of 46.4 degrees above the horizontal. How far from the base of the wall does the projectile land on the ground?(answer in m)

Next you launch another projectile form the top of the wall. After 5.5 seconds of flight, the projectile lands on the ground.

b) At what angle above the horizontal was the projectile launched this time? (answer in degrees)

c)How far from the base of the wall does the projectile land in this case? )answer in m)

Now you position the launcher on the ground, a distance of 41.2 m from the base of the wall, at an angle of 18.3 degrees above the horizontal.

d) how high above the ground does the projectile strike the wall?(answer in m)

e)Still positioned at the same distance 41.2 m from the base of the wall, what is the minimum launch angle with respect to the horizontal that will allow the prjectile to just make it over the top of the wall? ( answer in degrees)

f) what is the maximum distance from the bottom of the wall that you can position your launcher on the ground, such that its possible to make it over the wall? (answer in m)

Explanation / Answer

This is a widespread problem of missile launch

Data

Yo = 33.2 m

Vo =58.0 m /s

Part a) Kinematic equations used

Y = Yo + Voy t – ½ g t2(1)

= 46.4

We search the velocity components

Sin = Voy/Vo Voy = Vo Sin

Cos = vox/Vo Vox = Vo Cos

Voy = 58 Sin 46.4 Voy = 42 m/s

Vox = 58 Cos 46.4 Vox= 40 m/s

0 = 33.2 +42 t – ½ 9.8 t2

0= - 33.2 - 42 t + 4.9 t2

t= (42± sqrt (422 + 4 4.9 33.2) )/2 4.9

t =( 42±49,1) /9,8

t1 =( 42+ 49.1)/9.8

t1= 9,3 s

t2= (42-49.1) / 9.8

t2 = -0.72 s

we take the positive value

We estimate the range

X = Vox t (2)

X = 40 9,3

X =372 m

Then another missile launches and touches the ground after 5.5 s, consequently we can use the same expression

write the expressions 1 y 2

0= 33.2 + 58 Sin 5.5 - 4.9 5.52

X2 = 58 Cos 5.5

0 = 33.2 + 319 Sin -148.225

Sin = (148.225 – 33,2 )/319 = 0.36

=21.1

X2 = 319 Cos 21,1

X2 = 297.5 m

part b) Now we calculate it, the answer is = 21.1

part c) answre X2 = 297.5 m

Part d) and give us the horizontal distance (x3 = 41.2 m) calculated the time this distance and the height reached this value

3 = 18.3

X3 = Vo Cos 3 t

t = X3/(Vo Cos 3)

t =41.2 /( 58 cos 18.3)

t= 0.748 s

Y = Vo Sin 3 t - ½ g t2

Y = 58 Sin 18.3 0.748 - ½ 9.8 0.7482 = 13.62 -2.74

Y = 10.88 m

Part e) Y = 33.2 m

Part f) The maximum distance, this implies that the vertical velocity is zero

Vy2 = Voy2 – 2 g Y

Vy=0

Vo2 Sin2 5 = 2g Y

Sin5 = sqrt ( 2gY/Vo2)

Sin 5 = sqrt ( 2 9.8 33.2 / 582) = sqrt ( 0.1934)

5= 11.15

Now calculate the time to get to the wall

Vy = Voy -gt

Vy=0

t = Voy /g = Vo Sin 5 /g

t = 58 Sin 11.15 / 9.8

t = 1.144 s

the horizontal distance where is the pitcher

X5 = Vo Cos 5 t

X5 = 58 Cos 11.15 1.144

X5 = 65 m

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