A daredevil is shot out of a cannon at 45.0°.... please help!!!?

Question

A daredevil is shot out of a cannon at 45.0° to the horizontal with an initial speed of 26.0 m/s. A net is positioned a horizontal distance of 53.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the daredevil?

Answer 1

Initial responder apparently thinks the daredevil will travel in a straight line(!) That only happens if the circus takes place somewhere where there's no gravity. Let us assume that the circus is on earth. :-)

Consider the daredevil's horizontal and vertical velocities separately.

The daredevil's horizontal velocity is Vo(cosθ), where Vo is 26.0 m/s; and θ is 45°. The horizontal velocity remains constant.

Since the horizontal velocity is constant, you can figure out the time T it will take for him to cover a horizontal distance of 53.0 meters

T = d/v
T = (53.0m) / (horizontal velocity)
T = (53.0m) / (Vo(cosθ))

Now the question is: at what height will he be when T seconds have elapsed? For that, you need to use this equation which relates height, time, and initial _vertical_ velocity:

h = uT - gT²/2

In this equation, "u" is the initial vertical velocity, which is Vo(sinθ).

You now have all the numbers and equations you need: Just plug them in to find the value of "h".

Answer 2

assuming you are placing the net high in the air the height will also be 53.0 m. tan45= u/53 ----> u=53 since tan 45 = 1

Answer 3

The equation for vertical displacement for projectile motion is
y = x tan θ - (4.9) x^2 / u^2 (cos θ)^2.

Putting x = 53, θ = 45° and u = 26.0,

y = 53 tan 45° - (4.9)*(53)^2 / [(26)^2 * (cos 45°)^2]
= 53 - 40.7
= 12.3 m