A ball is thrown with an initial velocity of u=(10i +15j) m/s. Whan it reaches the top of it trajectory neglecting air resistanc
1 answer:
Answer:
v = (10 i ^ + 0j ^) m / s, a = (0i ^ - 9.8 j ^) m / s²
Explanation:
This is a missile throwing exercise.
On the x axis there is no acceleration so the velocity on the x axis is constant
v₀ₓ = 10 m / s
On the y-axis velocity is affected by the acceleration of gravity, let's use the equation
v_y = - g t
at the highest point of the trajectory the vertical speed must be zero
v_y = 0
therefore the velocity of the body is
v = (10 i ^ + 0j ^) m / s
the acceleration is
a = (0 i ^ - g j⁾
a = (0i ^ - 9.8 j ^) m / s²
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By Considering the vertical distance and both vertical and horizontal final velocity, the time t = 0.45 s and Velocity V = 6.7 m/s
Given that a Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table.
Height h = 1.0 m
As the ball rolls off the table, it will be fallen under gravity. Where
g = 9.8 m/
Initial vertical velocity = 0
Initial horizontal velocity = 5 m/s
Considering the vertical distance, the formula to use to calculate the time will be;
h = ut + 1/2g
1 = 0 + 1/2 x 9.8
1 = 4.9
= 1/4.9
t =
t = 0.45 seconds
It takes 0.45 seconds to hit the floor if no one sneezes.
To calculate its velocity when it hits the floor, we will need to calculate for both vertical and horizontal final velocity and find the resultant velocity of the two.
Vertical component
=
+ gt
= 0 + 9.8(0.45)
= 4.41 m/s
Horizontal component
=
+ at
but a = 0
= 5 m/s
Final velocity V =
V = 6.67 m/s
Therefore, it will hit the floor at a velocity of 6.7 m/s
Learn more here: brainly.com/question/5063616