An object is thrown up at 20m/s if it hits the ground after 10s,determine the final velocity and the height
1 answer:
1) The final velocity of the object is -78 m/s (downward)
2) The final height of the object is -290 m
Explanation:
1)
When the object is thrown up, it follows a free fall motion, subjected only to the force of gravity (that pulls the object downward). Therefore, the object has a constant acceleration
towards the ground (acceleration of gravity).
Since this is a motion with constant acceleration, we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a = g is the acceleration
t is the time of flight
In this problem we have:
u = +20 m/s (taking upward as positive direction)
t = 10 s is the time of flight
(downward, so negative)
Solving for v, we find the final velocity:
2)
For this part of the problem, we can use another suvat equation:
where
u is the initial velocity
a is the acceleration
t is the time
s is the displacement of the object
In this problem,
u = +20 m/s
t = 10 s
Substituting, we find the final position of the object with respect to the initial position:
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Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
