A car is brought to rest in a distance of 484m using a constant acceleration of -8.0m/s^2. What was the velocity of the car when
1 answer:
Answer:
88 m/s
Explanation:
To solve the problem, we can use the following SUVAT equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
d is the distance covered
For the car in this problem, we have
d = 484 m is the stopping distance
v = 0 is the final velocity
is the acceleration
Solving for u, we find the initial velocity:
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Hi there!
We can begin by calculating the time the ball takes to reach the highest point of its trajectory, which can be found using the following:
Where:
tmax = (? sec)
vsinθ = vertical comp. of velocity = 10sin(48) = 7.43 m/s)
g = acceleration due to gravity (9.8 m/s²)
We can solve for this time:
When the ball is at the TOP of its trajectory, its VERTICAL velocity is equivalent to 0 m/s. Thus, we can consider this a free-fall situation.
We must begin by solving for the maximum height reached by the ball using the equation:
d = displacement (m)
vi = initial velocity (7.43 m/s)
a = acceleration due to gravity
d = displacement (m)
y0 = initial VERTICAL displacement (28m)
Plug in the values:
Now, we can use the rearranged kinematic equation:
Add the two times together:
Answer:
The width of the river is
Explanation:
From the question we are told that
The distance of the base line is D = 130 m
The angle is
A diagram illustration the question is shown on the first uploaded image
Applying Trigonometric Rules for Right-angled Triangle,
Now making z the subject
