(a) The lunch angle is 64.8⁰
(b) The initial speed of the pass when the angle of projection is 25⁰ is 21.2 m/s
(c) The time of flight of the bullet is 1.83 s
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The given parameters include;
time of flight, T = 3.97 s
initial velocity, u = 21.5 m/s
(a) The lunch angle is calculated from the equation of motion of time of flight;
(b) the initial speed of the pass when the angle of projection is 25⁰ and range of 35 m, is calculated as follows;
(c) The time of flight of the bullet is calculated as;
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Answer:
4500 N
Explanation:
F = ma
= 1500 kg * 3 m/s^2 = 4500 N
edited to correct units !
Answer:
4.6x10^-7 m thickness
Explanation:
See attached handwritten document for more details
Answer:
Explanation:
Given
Distance between home and school is 39 km
Total time taken
i.e.
Total Distance traveled =39+39+39=117 km
displacement=39 km
Thus Average speed
Average velocity
Answer:
-10 m/s^2
Explanation:
Acceleration = Change of velocity over time
25/2.5 = 10 m/s^2
Since we are deccelerating here (Going slower) then the number is written as negative