a body is projected horizonatlly from the top of a cliff 45m above the ground. if the body lands at a distance 30m from the foot of the cliff, calculate the speed of projection. [g = 10ms-2]
10ms-1
15ms-1
20ms-1
30ms-1
Explanation
Video Explanation
No video available
Post your Contribution
Discussions (9)
The selected answer is based on vertical component of the motion and is the velocity with which the body strikes the ground.
The speed of projection does not change in the horizontal plane.
The right answer is A
Reasons:
1. Find v in vertical direction = 30 and use that to find T = 3
2. Use third equation of motion for the horizontal motion, with S = 30m, t= 3, g=0 (v=u for horizontal components of projectile motions)
U = s/t =30/3 =10m/s
To solve this, we will use the principles of projectile motion. Since the body is projected horizontally, its vertical and horizontal motions are independent of each other.
Step 1: Determine the time of flight
The time it takes for the body to fall vertically can be found using the equation:
ℎ
=
1
2
𝑔
𝑡
2
Where:
ℎ
=
45
m
(height of the cliff),
𝑔
=
10
m/s
2
(acceleration due to gravity),
𝑡
is the time of flight.
Rearranging for
𝑡
2
:
𝑡
2
=
2
ℎ
𝑔
Substitute the values:
𝑡
2
=
2
×
45
10
=
9
𝑡
=
9
=
3
s
Step 2: Calculate the horizontal velocity
The horizontal distance covered by the body is given by:
𝑥
=
𝑢
⋅
𝑡
Where:
𝑥
=
30
m
(horizontal distance),
𝑢
is the horizontal speed (speed of projection),
𝑡
=
3
s
(time of flight).
Rearranging for
𝑢
:
𝑢
=
𝑥
𝑡
Substitute the values:
𝑢
=
30
3
=
10
m/s
Final Answer:
The speed of projection is 10 m/s, which corresponds to option A.
