Motion in a plane: A projectile is fired with horizontal velocity of 330 m/s from the top of a 90m high tower. How long will it take to strike the level of the ground. With what velocity will it strike?
Answers 1
Step-by-step explanation:
This is assuming no air resistance, so the results will not be realistic.
For the range. S
S = (V cos 8)tflight where v is the launch velocity and tflight is the flight time To find the flight time, we need to find the time to reach the apex. v sin e t=
The flight time will be twice this time: 2v sin @ flight=2t=
Therefore, the range will be:
S₂ = (v cos 8) (2v sin 6) 9
2v2 cos e sin 0
v² sin(20)
65.49°
24.51
sin(20) gs =
sin-1(gs, 2 This gives you 2 possible angles: 6 and (90-0)
So, given the initial values:
sin-1
(32.17.52800)
15003
2
49.02
=24.51°
Since we want the maximum height, the angle will be 90° - 24.51 = 65.49° 1 The maximum height will be: h=gt where t will be the time to reach the apex
The time to reach the apex is: t= v sin 65.49 1500 (sin 65.49) g 32.17 42.43 seconds
So, the maximum height will be: h = (32.17)(42.43)2 = 28,958 ft
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