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 horizont
al 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?
2 answers:
Answer:
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
Explanation:
Answer:
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.
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