6. An aircraft is is travelling along a runway at a Velocity of 25m/s in It accelerates at a rate of 4m/s² for a distance of 750
1 answer:
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Answer:
d) None of the above
Explanation:
= inituial velocity of launch = 4 m/s
θ = angle of launch = 10 deg
Consider the motion along the vertical direction
= initial velocity along vertical direction = 4 Sin10 = 0.695
m/s
= acceleration along the vertical direction = - 9.8 m/s²
y = vertical displacement = - 20 m
t = time of travel
using the equation
- 20 = (0.695) t + (0.5) (- 9.8) t²
t = 2.1 sec
consider the motion along the horizontal direction
x = horizontal displacement
= initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s
= acceleration along the horizontal direction = 0 m/s²
t = time of travel = 2.1 s
Using the kinematics equation
x = (3.94) (2.1) + (0.5) (0) (2.1)²
x = 8.3 m
Consider the motion along the vertical direction
= initial velocity along vertical direction = 4 Sin10 = 0.695
m/s
= acceleration along the vertical direction = - 9.8 m/s²
=initial vertical position at the time of launch = 20 m
= vertical position at the maximum height = 20 m
= final velocity along vertical direction at highest point = 0 m/s
using the equation
= 20.02 m
h = height above the starting height
h = -
h = 20.02 - 20
h = 0.02 m
Answer:
The final position made with the vertical is 2.77 m.
Explanation:
Given;
initial velocity of the ball, V = 17 m/s
angle of projection, θ = 30⁰
time of motion, t = 1.3 s
The vertical component of the velocity is calculated as;
The final position made with the vertical (Yf) after 1.3 seconds is calculated as;
Therefore, the final position made with the vertical is 2.77 m.