A daredevil is shot out of a cannon at 45.0° to the horizontal with an initial speed of 31.0 m/s. A net is positioned a horizontal distance of 48.5 m from the cannon. At what height above the cannon should the net be placed in order to catch the daredevil?

s = vcos (x) t

50 = 25cos (45) t

cos (45) t = 2

t = 2/cos (45) = 2sqrt (2)

h = vsin (x) t + gt^2/2

h = 25sin (45) * 2sqrt (2) - 4.9*8

h = 10.8 metres

24.5 m

Explanation:

First, find the time it takes for the daredevil to travel 48.5 m horizontally:

x = x₀ + v₀ t + ½ at²

(48.5 m) = (0 m) + (31.0 m/s cos 45.0°) t + ½ (0 m/s²) t²

t = 2.21 s

Next, find the vertical position reached at this time:

y = y₀ + v₀ t + ½ at²

y = (0 m) + (31.0 m/s sin 45.0°) (2.21 s) + ½ (-9.8 m/s²) (2.21 s) ²

y = 24.5 m

The net should be placed 24.5 meters above the cannon.