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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Answer:
Part a)
Part b)
Direction = upwards
Explanation:
When ball is dropped from height h = 4.0 m
then the speed of the ball just before it will strike the ground is given as
Now ball will rebound to height h = 2.00 m
so the velocity of ball just after it will rebound is given as
Part a)
Average acceleration is given as
Part B)
As we know that ball rebounds upwards after collision while before collision it is moving downwards
So the direction of the acceleration is vertically upwards
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