Question

4. 2-D Kinematics An archer shoots an arrow with a velocity of 50.0m/s at an angle of 45 deg with the horizon An assistant standing on the level ground 150 m downrange from the launch point throws an apple straight up. g-9.8m/s a) What is the height of the arrow after it has traveled a horizontal distance of 150 m? b) What is the minimal launch speed of the apple to meet the path of the apple? c) At what time after the arrow launch should the apple be thrown so that the arrow hits the apple 5. 2-D Kinematics For a projectile that lands at the same height as it is released, show that the maximum horizontal distance is reached if the projectile is released at an angle of 45 deg with respect to the horizontal.

Answer 1

g = 9.8 m/s^2 heigh of arrow at 150 m 50 times cos 45 degree times t = 150 m [distance = speed times times] horizontal equation. t = 3 Squareroot(2) h = 50/squareroot (2) times t - 1/2 times 10 times 9 times 2 h = 50 t/Squareroot(2) times 1/2 times 10 times 9 times 2 h = 50 t/squareroot (2) - 88.2 h = 50/ Square (2) times 3 Squareroot(2) - 88 .2, h = 61.8 m minmum lauch speed is when maximum height of apple will be 61.8 m n_max = 61.8 m = v^2/2g = v = 34.80 m/s times is calculated in part (a) and comes out to be 3 Squareroot(25 = 4.23 sec. \r\n For Range of projectile we have formula R = v^2 sin 2 theta/g v rightarrow initial speed of the object theta rightarrow acceleration due to gravity For a given condition v and g will be same so only condition v and g will be