Question

Muhammad wishes to shoot an arrow through the open window of a tall building. The window is 32.8 meters above the ground and Muhammad stands 63.6 meters from the base of the building. If Muhammad aims the arrow at an angle of 51.5 degrees above the horizontal, with what minimum speed must he fire the arrow in order for it to enter the window?

Answer 1

Explanation:

Given Muhammad wishes to shoot an arrow through the open window of a tall building. The window is 32.8 meters above the ground and Muhammad stands 63.6 meters from the base of the building. If Muhammad aims the arrow at an angle of 51.5 degrees above the horizontal, with what minimum speed must he fire the arrow in order for it to enter the window?

• Now Vx = V1 cos theta and Vy = V1 sin theta.
• In the horizontal motion x = 63.6 m
• So Vx = V1 cos 51.5 degree
• Or Vx = V1 x 0.622
• Now for the vertical motion, y = 32.8 m
• So  Vy = V1 sin 51.5 degree
• Or Vy = V1 x 0.7826
• Also a x = 0 m/s and a y = 9.8 m/s
• Now for the horizontal displacement we have,
• So x = V1 t
• Or 63.6 = (0.6225 x V1) t --------------------1
• So for vertical displacement we have,
• So s = ut + ½ at^2
• Or y = Vy x t + ½ a y x t^2
•      32.8 = (0.7826 x V1) x t + ½ x (- 9.8) x t^2------------------2
• From equation 1 we get  
• So t = 63.6 / 0.6225 x V1
• Substituting this in equation 2 we get
•   32.8 = (0.7826 x V1) x (63.6 / 0.6225 x V1) + ½ (- 9.8) x (63.6 / 0.6225 x V1)^2
•   32.8 = 79.957 – 51145.94 / V1^2
•  So – 47.157 = - 51145.94 / V1^2
• Or V1^2 = - 51145.94 / - 47.157
• Or V1^2 = 1084.58
• Or V1 = 32.9 m/s

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https://brainly.in/question/11902935