When an object threw to the free space to make an angle of 25 degree at an initial speed of 15 m/sec, the ball takes time to rea
1 answer:
Answer:
The horizontal distance traveled by the projectile is 15.23 m.
Explanation:
Given;
angle of projection, θ = 25⁰
initial velocity of the projectile, u = 15 m/s
time of flight, t = 1.12 s
The the travelling path of the object is calculated as the range of the projectile
Therefore, the horizontal distance traveled by the projectile is 15.23 m.
You might be interested in
Answer:
Check attachment for better understanding
Explanation:
Given that
a= 0.75m
b=1.31m
c= 2.2m
Weight of pole is 26.90
Then, Fg = Weight = 26.90
Using Equilibrium of forces
ΣFy = 0
U — D — Fg = 0
U — D = Fg
U — D = 26.9
To calculate U,
We will take moment about point A.
ΣMa = 0
Let the clockwise moment be positive and anti-clockwise be negative
Fg(a+b) — U(a) = 0
26.9(1.31+0.75) —0.75U = 0
26.9(2.06) = 0.75U
0.75U = 55.414
U = 55.414/0.75
U = 73.89 N
To calculate D,
U — D = 26.9
73.89—D =26.9
73.89—26.9 = D
D = 46.99N
Answer:
(A) –14m/s
(B) –42.0m
Explanation:
The complete solution can be found in the attachment below.
This involves the knowledge of motion under the action of gravity.
Check below for the full solution to the problem.
