an object is launched from a cliff 20 m above the ground at an angle of 30 degrees above the horizontal with an initial speed of 30m/s how far does the object travel before landing on ground (in metres)
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Explanation:
Height of cliff = 20 m
Initial speed = 30 m/s & angle = 30°
Horizontal speed = 30Cos30° = 15√3 m/s
Vertical Speed = 30Sin30° = 15 m/s
Vertical Speed at top = 0
0 = 15 - gT => 0 = 15 - 10T => 1.5 sec
Vertical Distance in 1.5 sec = 15(1.5) - (1/2)(10)(1.5)² = 22.5 - 11.25 = 11.25 m
Horizontal Distance = 1.5 * 15√3 = 22.5√3 m
Height at top = 20 + 11.25 = 31.25 m
31.25 = (1/2)(10)T²
=> T = 2.5 sec
Vertical Distance = 31.25 m
Horizontal Distance = 15√3 * 2.5 = 37.5√3 m
Total Horizontal Distance covered = 22.5√3 + 37.5√3 = 60√3 m
Total Vertical Distance covered = 11.25 + 31.25 = 42.5 m
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