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

Question

You are outside in a big level field in the midst of which stands a narrow wall, 25 m high. you are given a device that can launch a projectile, always with the same speed of 40m/s.

a.) First you launch a projectile from the top of he wall at an angle of 55 degrees above horizontal. how far from the base of the walll does the projectile land on the ground?

Next you launch another projectile from the wall. After 5 seconds of flight the projectile lands on the ground.

b.) at what angle was the projectile launched this time?

c.) how far from the base of the wall does the projectile land in this case?

Next the launcher is positioned on the ground a distance of 50m from the base of the wall. You launch the projectile at the wall with an angle of 21 degrees.

d.) how high above the ground does the projectile strike the wall?

e.) still at the same distance 50m from the base of the wall what is the minimum launch angle with respect to the horizontal that will allow the projectile to just make it oover the top of the wall?

f.) What is the maximum distance from the wall that you can launch from and still be able to make it over the top of the wall?

I really need to undertsand how this is done in order to prepare for my exam. A drawn out explanation woud be greatly appreciated thank you in advance.

Explanation / Answer

a) y direction
y = y0 + v0y t + 1/2 a t^2
0 = 25 + 40*sin(55)*t - 0.5*9.81*t^2
t=7.37 s

x direction

x = 40*cos(55)*7.37= 169 m

b)

so now
0 = 25 + 40*sin(theta)*5 - 0.5*9.81*5^2
sin(theta) = 0.488
theta = 29.21 degrees

c) x = 40*cos(29.21 degrees)*5= 174.6 m

d) x direction

50 = 40*cos(21 degrees)*t
t=1.34 s

y = y0 + v0y t + 1/2 a t^2 = 40*sin(21 degrees)*1.34 - 0.5*9.81*1.34^2= 10.4 m

e)
y direction

v^2 = v0^2 + 2 a y
0 = (40*sin(theta))^2 - 2*9.81*25
sin theta = 0.554

theta = arcsin(0.554)= 33.64 degrees

f) need to find the time to just barely make to top of the wall
y = y0 + v0y t + 1/2 a t^2
25 = 0 + 40*sin(33.64)*t - 0.5*9.81*t^2
t=2.19 s

so x = 40*cos(33.64 degrees)*2.19= 73 m

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