V 0 cos(35°) = 30 √ Ī ball kicked from ground level at an initial velocity of 60 m/s and an angle θ with ground reaches a horizontal distance of 200 meters. √ < V 0 < √ Ī ball is kicked at an angle of 35° with the ground.Ī) What should be the initial velocity of the ball so that it hits a target that is 30 meters away at a height of 1.8 meters?ī) What is the time for the ball to reach the target?ġ.8 = -(1/2) 9.8 (30 / V 0 cos(35°)) 2 + V 0 sin(35°)(30 / V 0 cos(35°)) We want to have the range greater than OM and smaller that ON, The range is given by x = V 0 2 sin(2θ) / g OM (t=1.16)= √ = 15 metersĪ projectile is to be launched at an angle of 30° so that it falls beyond the pond of length 20 meters as shown in the figure.Ī) What is the range of values of the initial velocity so that the projectile falls between points M and N? The x and y components of the displacement are given by The projectile hits the incline plane at point M.Ī) Find the time it takes for the projectile to hit the incline plane. √ = V0 = 20 m/sĪ projectile is launched from point O at an angle of 22° with an initial velocity of 15 m/s up an incline plane that makes an angle of 10° with the horizontal. The magnitude V of the velocity is given by The components of the velocity at t = 2 V0 sin(θ) / g are given by The components of the velocity at t are given by ![]() The object hits the ground at t = t2 = 2 V0 sin(θ) / g (found in part b above) The horizontal range is the horizontal distance given by x at t = t2. Time of flight = t2 - t1 = 2 (20) sin(θ) / g = 1.72 seconds.Ĭ) In part c) above we found the time of flight t2 = 2 V 0 sin(θ) / g. Maximum height y (0.86) = 20 sin(25°)(0.86) - (1/2) (9.8) (0.86) 2 = 3.64 metersī) The time of flight is the interval of time between when projectile is launched: t1 and when the projectile touches the ground: t2. T = V 0 sin(θ) / g = 20 sin(25°) / 9.8 = 0.86 secondsįind the maximum height by substituting t by 0.86 seconds in the formula for y That is when the projectile changes from moving upward to moving downward.(see figure above) and also the animation of the projectile. ![]() The height of the projectile is given by the component y, and it reaches its maximum value when the component V y is equal to zero. Detailed SolutionsĪn object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.Ī) What is the maximum height reached by the object?ī) What is the total flight time (between launch and touching the ground) of the object?Ĭ) What is the horizontal range (maximum x above ground) of the object?ĭ) What is the magnitude of the velocity of the object just before it hits the ground?Ī) The formulas for the components V x and V y of the velocity and components x and y of the displacement are given by These solutions may be better understood when Solutions and detailed explanations to projectile problems are presented.
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