Answer:
The other angle is 75⁰
Explanation:
Given;
velocity of the projectile, v = 10 m/s
range of the projectile, R = 5.1 m
angle of projection, 15⁰
The range of a projectile is given as;
To find another angle of projection to give the same range;
<u>Check: </u>
sin(2θ) = sin(2 x 75) = sin(150) = 0.5
sin(2θ) = sin(2 x 15) = sin(30) = 0.5
Answer:
Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s
Explanation:
Given Data:
V is the overall velocity vector, and
are its initial vertical and horizontal components
To find:
Max Height achieved
Calculation:
1) Using the equation of motion, we know
2) In terms of gravity height
and the vertical component of Velocity
.
3) As as at maximum height the vertical component of velocity is zero maximum height achieved
putting values
4)
5) As for the speed when it reaches back its starting point, it will have a speed similar to its launching speed, the reason being the absence of air friction (Air drag)
Explanation:
The work done to lift the bag of weight mg through height 2.5 m is 400 J
The work done can be found by relation W = mg x h
Thus mg = =
= 160 N
Therefore the man have to apply the force of 160 N