A soccer ball is kicked from the ground with an initial speed of 12m/s at an angle of 32 degrees above the horizontal. What are the x and y positions of the ball 0.5s after it is kicked? What are the x and y components for the balls velocity?
Using y = sen(32) × 12 and x = cos(32) × 12X and Y components:Y = 6.36 m/sX = 10.18 m/sPosition after 0.5 seconds:Y = 0.5 × 6.36 m/s = 3.18 metersX = 0.5 × 10.18 m/s = 5.09 metersPosition(5.09, 3.18)
A soccer ball is kicked from the ground with an initial speed of 18.6 m/s at an upward angle of 45°?
Break the 18.6 into x and y vector quantities like a triangle. 18.6 is the hypotenuse and x and y are the bases. cos(45)= x/18.6 x= 13.13m sin(45)=y/18.6 y=13.13m To find the time the ball is in the air, you must first find how high the ball goes. Y(velocity) = Y(initial velocity) - gt Y(velocity) equals zero at the top of the arc and you now know Y(initial velocity) so do this 0=13.13 -(9.8)t t=1.34s This only gets you to the top of the arc because that is where velocity changes from positive to negative. So to get the total time, multiply that by two. 1.34 x 2 = 2.68s To find how far it moved in the x direction, multiply time by velocity 2.68 x 13.13 = 35.19m So the ball will travel 35.19 meters in 2.68 seconds. That means the other player has to cover (56.4-35.19) so 21.21 meters in 2.68 seconds. Divide distance by time and you get 7.91 m/s he has to run. That answer makes sense since that is a possible speed to run. Drawing a picture usually helps too. Just try to imagine everything in terms of x and y distance and times instead of angles. That's what we did in the first step. Hope this helps!
A soccer ball is kicked from the ground with an initial speed v at an upward angle θ. A player a distance d aw?
yeah me neither...physucks 2a ucsd? hahhaha
Physics Help!!!A soccer ball is kicked from the ground with an initial speed of 19.6 m/s at an upward angle of 47.5˚....?
A soccer ball is kicked with an initial horizontal velocity of 19 m/s and an initial vertical velocity of 14 m?
What is the initial speed of the ball? What is the initial angle θ of the ball with respect to the ground? What is the maximum height the ball goes above the ground? How far from where it was kicked will the ball land? What is the speed of the ball 1.3 seconds after it was kicked? How high above the ground is the ball 1.3 seconds after it is kicked? somebody please help. thanks
A soccer ball is kicked with an initial horizontal velocity of 13.0 m/s and an initial vertical velocity of 19?
A soccer ball is kicked with an initial horizontal velocity of 13.0 m/s and an initial vertical velocity of 19.0 m/s. Answer the following: 1) What is the initial speed of the ball? 2) What is the initial angle $\theta$ of the ball with respect to the ground? 3) What is the maximum height the ball goes above the ground? 4) How far from where it was kicked will the ball land? 5) What is the speed of the ball 1.4 seconds after it was kicked? 6) How high above the ground is the ball 1.4 seconds after it is kicked?
A soccer ball is kicked with an initial horizontal velocity of 11 m/s and an initial vertical velocity of 15 m/s.?
i) What is the initial speed of the ball? v = √[11² + 15²] = 18.601 m/s ii) What is the initial angle θ of the ball with respect to the ground? initial θ = tan ̄ ¹(15/11) = 53.75° iii) What is the maximum height the ball goes above the ground? H = [u(vertical)]² / 2g = 15² / 19.62 = 11.47 m iv) How far from where it was kicked will the ball land? R = horizontal u x t t = time of flight = 2[u(vertical)] / g = 30 / 9.81 = 3.058 s => R = 11 x 3.058 = 33.64 m 5) What is the speed of the ball 1.7 seconds after it was kicked? after 1.7 s , vertically v = u + at = 15 - (9.81)1.7 = -1.677(downwards) therefore since horiz. speed = 11 m/s is constant, resultant speed = √[11² + 1.677²] = 11.13 m/s 6) How high above the ground is the ball 1.7 seconds after it is kicked? s = vₐᵥₑt = (-1.677 + 15)1.7/2 = 13.323(1.7)/2 ~= 11.33 m above the ground(while descending) hope this helps
A soccer ball is kicked from the ground at an angle of 340 with the horizontal at an initial speed of 40 ft/s.?
A soccer ball is kicked from the ground at an angle of 340 with the horizontal at an initial speed of 40 ft/s. Ignoring air friction, determine: a.) how long before it hits the ground b.) how high it will go c.) how far it will go
A soccer ball, kicked from ground level at an angle of 72.0° above horizontal, is in the air for 3.30 s.?
It rises to maximum height in (3.3/2) = 1.65 secs. Its initial vertical V component = (gt) = 9.8 x 1.65 = 16.17m/sec. Its initial speed from kickoff = (16.17/sin 72) = 17m/sec.
A soccer ball is kicked from the ground...?
with an initial speed of 19.5m/s at an upward angle of 45 degrees. A player 55m away in the direction of the of the kick starts running to meet the ball at that instant. What must his average speed be if he is to meet the ball just before it hits the ground? I actually know how to do the majority of this problem. I'm just wondering if the very last line affects the answer at all? If I want to know his average speed just before it hits the ground, would I have to account for and alter the time or distance in the equation so it's just before it hits the ground instead of exactly when it hits the ground???