Answer:
Explanation:
First we need to determine the distance covered during deceleration. According to the equation of motion.
S = ut+1/2at²
Given:
u = 20m/s
t = 0.50s
a = -10m/s (deceleration is negative acceleration)
S = 20²+1/2(-10)(0.5)²
S = 400-5(0.5)²
S = 400-5(0.25)
S = 400-1.25
S = 398.75m
If the deer steps onto the road 35m in front of you, the distance between you and the deer when you come to a stop will be 398.75-35 = 363.75m
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
1,3 and 5 are the answers
a una velocidad de
22 m/s, quien lo golpea y devuelve en la misma
dirección con una velocidad de 14 m/s. Si el
tiempo de contacto del balón con la jugadora es
de 0,03 s, ¿con qué fuerza golpeó la jugadora el
balón?
21 Una bala de 0,8 g, está en la recámara de un rifl e
cuando se g
