A ball is to be shot from level ground toward a wall at distance x (see the figure (a)). The figure (b) shows the y component vy of the ball's velocity just as it would reach the wall, as a function of that distance x. The scaling is set by vys = 5.0 m/s and x= 20 m. What is launch angle? (a) x(m) (b) (s/u) 4
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- what’s the initial and final height of the ball?In the absence of air resistance, a projectile is launched from and returns to ground level. It has a range of 16m. Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?A stunt car traveling at 25 m/s flies horizontally off a cliff and lands 65.8 m from the base of the cliff. a) Sketch the scenario. b) Determine how tall is the cliff?
- please i need the full answer with the stepsA parent at a Little League game returns a foul ball by throwing it back onto the field. When the parent throws the ball back, she is standing and ends up releasing the ball from a point that is 2.50m above the ground. Your best guess is that she threw the ball upward at an angle of 60o with respect to the ground and it landed 13m away from her. What was the height of the ball's apex (as measured from the ground)?The initial speed of a tennis ball is 57.5 m/s and the launch angle is ?i = 16°. Neglect air resistance.Question 1) What is the maximum height, h, of the tennis ball? in mQuestione 2) What is the range, R, of the tennis ball? in m info given: what angle results in the greatest height=90 For this 90° angle, the horizontal component of velocity is zero, so the initial velocity is completely in the vertical direction, resulting in greatest height Kato tries substituting ty,max for t, 0 for yi, and h for yf, and gets h = vi2 sin2 ?i 2g When ?i = 90°, sin2 ?i is maximum, so h is maximum what angle results in the maximum horizontal range=45° I substituted 2vi sin ?i g into the expression for the horizontal component of velocity, (vi cos ?i)t, and got R = 2vi2 sin ?i cos ?i g when ?i = 45°, 2?i = 90°, and sin 90° = 1, which is its maximum value.
- This problem will involve deriving a formula or two for a projectile launched from one height and angle and landing at a different height on Earth. Begin with a projectile launched at angle 0 above horizontal from a height y₁ with initial velocity Vo. The projectile lands at a point with height y₂. These are the given quantities: vo, 0, y₁, y2 and g. Construct formulae for each of the following, as. a function of given quantities the horizontal distance traveled. the maximum height reached. the time taken. the angle of impact. (find the final velocity components first).= 3.98 m to the north, A novice golfer on the green takes three strokes to sink the ball. The successive displacements of the ball are d₁ d₂ = 2.06 m northeast, and d3 = 1.16 m at 30.0° west of south (see the figure below). Starting at the same initial point, an expert golfer could make the hole in what single displacement? magnitude direction WE S m o north of east d₂ d₁ 30.0° dzIT.0.0a Given an acceleration vector, initial velocity (uo.Vo), and initial position {xo.yo}, find the velocity and position vectors for t2 0. a(t) = (cos t,5 sin t. (Uo Vo) = (0,5) , (Xo.Yo) = (2.0) What is the velc | vector? v(t) =