C CrzySxyQwel New member May 7, 2013 #1 ...90ft, find the initial speed of the bal? If Roger hit the ball at 45deg, 1ft from the ground, and it traveled 90ft, find the initial speed of the ball in m/s.
...90ft, find the initial speed of the bal? If Roger hit the ball at 45deg, 1ft from the ground, and it traveled 90ft, find the initial speed of the ball in m/s.
S Saurabh_IND New member May 7, 2013 #2 The initial velocity of an object is speed of the object before the application of the force. As we know that force is defined as f = m×a where, m = mass of the object; and a = acceleration of the object. So, the final velocity is defined as: v(f) = v(i) + a×t where, v(f) = final velocity of the object; v(i) = initial velocity of the object; and a = acceleration of the object in the given time t.
The initial velocity of an object is speed of the object before the application of the force. As we know that force is defined as f = m×a where, m = mass of the object; and a = acceleration of the object. So, the final velocity is defined as: v(f) = v(i) + a×t where, v(f) = final velocity of the object; v(i) = initial velocity of the object; and a = acceleration of the object in the given time t.
O oldprof New member May 8, 2013 #3 From x(T) = Ux T = 90 ft. we have T = 90/Ux as the total flight time. Then y(T) = h + Uy T - 1/2 g T^2 = h + Uy (90/Ux) - 1/2 32.2 (90/Ux)^2 = 0 = 1 + tan(45) 90 - 16.1 * 90^2/Ux^2 is the height at any time t = T. So 1 + tan(45) 90 = 16.1 * 90^2/Ux^2 and Ux^2 (1 + tan(45) 90) = 16.1*90^2 And then Ux = sqrt(16.1*90^2/(1 + tan(radians(45))*90)) = 37.85600247 = 37.86 ft/s. And finally U = Ux/cos(45) = 37.86/cos(radians(45)) = 53.54212547 = 53.5 ft/s. ANS. NOTE: Disregard radians, that's for Excel to work.
From x(T) = Ux T = 90 ft. we have T = 90/Ux as the total flight time. Then y(T) = h + Uy T - 1/2 g T^2 = h + Uy (90/Ux) - 1/2 32.2 (90/Ux)^2 = 0 = 1 + tan(45) 90 - 16.1 * 90^2/Ux^2 is the height at any time t = T. So 1 + tan(45) 90 = 16.1 * 90^2/Ux^2 and Ux^2 (1 + tan(45) 90) = 16.1*90^2 And then Ux = sqrt(16.1*90^2/(1 + tan(radians(45))*90)) = 37.85600247 = 37.86 ft/s. And finally U = Ux/cos(45) = 37.86/cos(radians(45)) = 53.54212547 = 53.5 ft/s. ANS. NOTE: Disregard radians, that's for Excel to work.
P PalashBansal New member May 13, 2013 #4 it is a projectile question solve it through Newton's equation let v be the initial velocity In horizontal direction s = ut + 1/2 at^2 here a = 0 90 = v * cos45 * t + 0 t = 90*root 2 /v In vertical direction s = ut + 1/2 at^2 1 = v*sin45 * 90*root2/v - 1/2 * 9.8 * 90*90*2 / v^2 v = root( 4.9 * 8100 / 89 ) v = 21.12 ft/s v = 7 m/s * are you sure about the values
it is a projectile question solve it through Newton's equation let v be the initial velocity In horizontal direction s = ut + 1/2 at^2 here a = 0 90 = v * cos45 * t + 0 t = 90*root 2 /v In vertical direction s = ut + 1/2 at^2 1 = v*sin45 * 90*root2/v - 1/2 * 9.8 * 90*90*2 / v^2 v = root( 4.9 * 8100 / 89 ) v = 21.12 ft/s v = 7 m/s * are you sure about the values