Since momentum is a vector quantity, take any direction as positive and other as negative. Answer won't change.
water balloons are tossed from the roof of a building, all with the same speed but with different launch angles. Which one has t
1 answer:
Answer:
90°/-90°
Explanation:
assume we are measuring launch angles with respect to horizontal surface
assume +y direction to be positive
when object is launched with launch angle <90° but >0°
it has 2 components for velocity v
in horizontal direction
in vertical direction
the <u>component remains unchanged throughout the travel</u> as there is no force acting on it to change it (air resistance is ignored)
this component is decreased till 0, as acceleration due to gravity is acting on it and<u> starts to increase in downward direction</u>.
when ball reaches the same height of the building while going down, this vertical component of ball is equal same as but in opposite direction
so now vertical velocity = -
this <u>component keeps on increasing until it hits the ground</u> as the gravity accelerates it to the ground
the horizontal component is unchanged throughout
So to have maximum speed when the ball reaches the ground,the vertical velocity component must be maximum at launch,so it will <u>just be accelerated until it hits the ground.</u>
⇒ In this case,the maximum vertical component can be v or -v
⇒=v or
= -v
⇒ horizontal component is 0
this means that ball is thrown directly up at 90° or directly down at 90° with respect to horizontal.
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1)
The initial data of the projectile are:
is the initial height
is the initial speed of the projectile, so its components along the x- and y- directions are
The motion of the cannonball along the x-direction is a uniform motion with constant speed, while on the y-direction it is an uniformly accelerated motion with constant acceleration g=9.8 m/s^2 downward. So, the two equations of motion of the projectile along the two directions are:
2) 4.94 s
To determine how long the cannon ball was in the air, we need to find the time t at which the cannonball hits the ground, so the time t at which y(t)=0:
Solving the equation with the formula, we have:
which has two solutions:
t = -0.50 s
t = 4.94 s
Discarding the first solution which is a negative time so it has no physical meaning, the correct solution is
t = 4.94 s
3) 115.6 m
To determine how far the cannonball travelled, we need to find the value of the horizontal position x(t) when the ball hits the ground, at t=4.94 s. Substituting this value into the equation of motion along x, we find:
4) 2.22 s
The cannonball reaches its maximum height when the vertical velocity becaomes zero.
The vertical velocity at time t is given by
where
g = 9.8 m/s^2 is the acceleration due to gravity
Substutiting and solving for t, we find
5) 36.2 m
The maximum height reached by the cannon is equal to the vertical postion y(t) when the vertical velocity is zero, so when t=2.22 s. Substituting this value into the equation of the vertical motion, we find: