An object is launched from a cliff 20 m above the ground at an angle of 30° above the horizontal with an initial speed of 30 m/s. How far horizontally does the object travel before landing on the ground. (g = 10 m/s²)
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# Answer- 104 m
## Explaination-
# Given-
u = 30 m/s
h = 20 m
θ = 30°
# Solution-
When the body is thrown above, constant gravitational accn acts on it.
Vertical velocity
uy = usinθ
uy = 30×sin30
uy = 15 m/s
Time of ascent
t1 = uy/g
t1 = 15/10 = 1.5 s
Height above cliff
H = u^2/2g
H = (15)^2/(2×10)
H = 11.25 m
Total height
s = H+h
s = 20+11+25
s = 31.25 m
Time of descent
t2 = √(2s/g)
t2 = √(2×31.25/10)
t2 = 2.5 s
Total time
t = t1+t2
t = 1.5+2.5
t = 4 s
Total horizontal distance
l = utcosθ
l = 30×4×cos30
l = 60√3
l = 103.9 m
Total horizontal distance travelled by body is 104 m.
Hope this is helpful..
# Answer- 104 m
## Explaination-
# Given-
u = 30 m/s
h = 20 m
θ = 30°
# Solution-
When the body is thrown above, constant gravitational accn acts on it.
Vertical velocity
uy = usinθ
uy = 30×sin30
uy = 15 m/s
Time of ascent
t1 = uy/g
t1 = 15/10 = 1.5 s
Height above cliff
H = u^2/2g
H = (15)^2/(2×10)
H = 11.25 m
Total height
s = H+h
s = 20+11+25
s = 31.25 m
Time of descent
t2 = √(2s/g)
t2 = √(2×31.25/10)
t2 = 2.5 s
Total time
t = t1+t2
t = 1.5+2.5
t = 4 s
Total horizontal distance
l = utcosθ
l = 30×4×cos30
l = 60√3
l = 103.9 m
Total horizontal distance travelled by body is 104 m.
Hope this is helpful..
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