A soccer player heads the ball and sends it flying vertically upwards at a speed of 18.0 m/s . How high above the players ' head
1 answer:
Answer:
<em>The ball travels up to 16.53 meters above the player's head</em>
Explanation:
<u>Vertical Launch Upwards</u>
In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.
If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:
The initial speed of the soccer ball is vo=18 m/s. The initial height can be assumed to be zero because we are required to find the maximum height with respect to the player's head, where the vertical motion was initiated.
Calculate the maximum height:
The ball travels up to 16.53 meters above the player's head
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Answer:
(a) The angle of projection is 63 degree.
(b) The velocity of projection is 24.5 m/s.
Explanation:
Height, h = 1 m
horizontal distance, d = 50 m
time, t = 4.5 s
Let the initial velocity is u and the angle is A.
(a) Horizontal distance = horizontal velocity x time
50 = u cos A x 4.5
u cos A = 11.1 .....(1)
Use second equation of motion in vertical direction
Divide (2) by (1)
tan A = 1.97
A = 63 degree
(b) Substitute the value of A in equation (2)
u x sin 63 = 21.8
u = 24.5 m/s
Answer:
a)
Explanation:
- Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:
- Since the car starts from rest, v₀ =0.
- We know the value of t = 5 sec., but we need to find the value of a.
- Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:
- Replacing a and t in (1):
- Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
- Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:
- Replacing v₀, at and t in (1), we have:
- Therefore, as the truck travels twice as far as the car, the right answer is a).