Answer:
v₀ₓ = 28.6 m / s
Explanation:
This is a missile throwing exercise
y = 
t - ½ g t²
as the dart is thrown horizontally the vertical velocity is zero (I go = 0)
y = - ½ g t²
t = 
let's calculate
t = 
t = 0.175 s
the expression for horizontal displacement is
x = v₀ₓ t
v₀ₓ = x / t
v₀ₓ = 5.00 / 0.175
v₀ₓ = 28.6 m / s
Answer:
The maximum height reached by the ball is 2.84 m
Explanation:
Given;
initial velocity of the soccer, u = 13 m/s
angle of projection, θ = 35°
The maximum height reached by the ball = ?
Apply the following kinematic equation, to determine the maximum height reached by the ball.
Maximum height (H) is given as;
Therefore, the maximum height reached by the ball is 2.84 m
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Answer:
The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.
Explanation:
1200 times 2.5 is 3000. So it flew 3000 kilometers.
