Dont mind the top part, hope this helped
Answer:
the maximum height reached by the object from where it was launched is 0.4591 m
Explanation:
initial speed of the object, u= 3 m/s
The velocity at the maximum height will always be 0.
Therefore, final velocity, v= 0 m/s
Using the Newton's equation of motion,
v^2 - u^2 = 2*g*h(max)
0 - u^2= 2*g*h(max)
h(max) = -u^2 /2* g
where g is the gravitational acceleration.
g= - 9.8 m/s^2
substituting the values in equation,
h(max)= - (3*3) / 2*(-9.8)
h(max) = 0.4591 m
the maximum height reached by the object from where it was launched is 0.4591 m
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Answer:
The only force acts on a projectile is gravitational force {F<em>g</em>}, therefore its acceleration a=F<em>g</em>/m will always directed towards the direction of force i.e. vertically downwards. Therefore it will always be perpendicular to the x direction or here we can say that a is always perpendicular to Vx}.
Explanation:
The vector r indicates the instantaneous displacement of a projectile from the origin. At the instant when the projectile is at r , its velocity and acceleration vectors are v and a . Which statement is correct?