An object is projected so that it just clears two walls of
height 7.5 m and with separation 50m from each other.
If the time passing between the walls is 2.5s,the range of the projectile will be ?(g= 10 m/s)
Answers
Answer:
h = 750 m
t is the time taken to travel vertical distance of 750m
750 = usinθ t−1/2gt^2
t^2−2usinθ/g+1500/g=0
roots of this equation will be t1 and t2
where a1 and a2 are the timing to the 1at and second wall
a1 a2 = 1500/10 = 150
and a1−a2 = 19 given
a1 = 25 s
distance travel between the plates
60= ucosθ*19
ucosθ = 60/19
r is the horizontal distance tarvelled while reaching the first wall
r = ucosθa1 = 60/19×25 = 78.94
R = 2r+60 = 217.94 m
Hence the range of the projectile will be 217.94 m
Answer:
70m
Explanation:
Horizontal component of projectile's motion
2.5
50
=20㎧
To have a range of 50m, maximum height would be reached in
2
2.5
=1.25secs (neglecting walls for now)
1.25×g=initial velocity=12.5㎧
Height attained over walls=
2g
V
2
=
2g
(12.5)
2
=7.8125m
Adding wall weight(7.812+7.5)=15.3125m
above ground level
So initial vertical v component from ground level
=
2gh
=
2×10×15.3125
=17.5㎧
Time to maximum height from the ground=
g
V
=
10
17.5
=1.75sec
Total time in air=(1.75×2)=3.5sec
Range=3.5×20
=70m
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