Answer: 4 s
Explanation:
Given
The ball leaves the hand of student with a speed of 
When the hand is
above the ground
Using the equation of motion we can write

Substitute the values
![\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]](https://tex.z-dn.net/?f=%5CRightarrow%202.5%3D-19t%2B0.5%5Ctimes%209.8t%5E2%5C%5C%5CRightarrow%204.9t%5E2-19t-2.5%3D0%5C%5C%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B19%5Cpm%20%5Csqrt%7B%28-19%29%5E2-4%5Ctimes%204.9%5Ctimes%20%28-2.5%29%7D%7D%7B2%5Ctimes%2019%7D%5C%5C%5CRightarrow%20t%3D4.0049%5Cquad%20%5B%5Ctext%7BNeglecting%20the%20negative%20value%20of%20%7Dt%5D)
Thus, the ball will take 4 s to hit the ground.
Answer:
Vy = 26 m/s sin 30 = 13 m/s vertical speed
t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec time to reach Vy = 0
H = Vy t + 1/2 g t^2
H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m
Answer:
The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>
Explanation:
Given:
Upward direction is positive. So, downward direction is negative.
Tota time the ball remains in air (t) = 8.0 s
Net displacement of the ball (S) = Final position - Initial position = 0 m
Acceleration of the ball is due to gravity. So,
(Acting down)
Now, let the initial velocity be 'u' m/s.
From Newton's equation of motion, we have:

Plug in the given values and solve for 'u'. This gives,

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.
The yellow light indicates that you have to slow down and slowly come to a stop. You slow your car until light completely turns red, then you stop at red and wait for the light