At what velocity should he throw the bag and how far from the house is he standing when he throws the bag in the following case?A boy wants to throw a bag of candy into the open window of his...
At what velocity should he throw the bag and how far from the house is he standing when he throws the bag in the following case?
A boy wants to throw a bag of candy into the open window of his girlfriend's bedroom 10 m above. It just reaches the window when he throws the love gift at 60.0 degree to the ground.
I think justaguide meant to use V*sin60 in the first calculation so all the values are vertical ones.
At the top of the trajectory when it "just" gets into the window it will have 0 vertical velocity (Vf) and had an initial vertical velocity (Vi) = V*sin60.
Using Vf^2 - Vi^2 = 2*a*d
=> 0 - (V*sin60)^2 = 2*-9.8*10
minus signs drop from both sides and divide by (sin60)^2 =.75
=> V^2 = 2*9.8*10/.75 V^2 = 261.3
so V = 16.16 m/sec
Using a = (Vf-Vi)/t solve for t to get t = (Vf-Vi)/a
and again using only vertical values/components
=> t = (0 - 16.16*(sin60)) /-9.8
it takes 1.428 seconds to reach the window.
so now using only the horizontal component of the throwing velocity V*cos60 you can use the Vavg = d/t to find the horizontal distance he is standing from the window.
V = d/t solve for d
d = V * t = 16.16(cos60) * 1.428 = 11.5 m
he is standing 11.5 m from the window.
Generally with trajectories the peak hieght of the trajectory is about half the height it would reach if there was no gravity at all. So if he threw it in a straigh line (no gravity) it would hit a window 20 m high if he was standing 11.5 m away. if you draw the diagram you will see that 20 over 11.5 is very close to the tangent of 60 degrees which is a quick check to see if the answer is in the ballpark (or window as the case may be)
The boy wants to throw a bag of candy into the open window of his girl-friend's bedroom 10 m above. It reaches the window when he throws it at an angle of 60 degrees to the ground. The velocity at which the bag is thrown and the distance from the house he is standing at has to be determined.
Let the velocity at which it is thrown be V.
The vertical component of the velocity is V*sin 60 and the horizontal component is V*cos 60.
The vertical acceleration on the bag is 9.8 m/s^2 acting vertically downwards. When the bag reaches the window, the vertical velocity is 0. 0
Use the relation v^2 - u^2 = 2as.
0 - (V*sin 60)^2 = -2*9.8*10
=> V^2*0.75 = 2*9.8*10
=> V^2 = 261.33
=> V = 16.16 m/s
The bag is thrown at 16.16 m/s
The time taken by the bag to travel the 10 m is (16.16*sin 60/9.8)
The horizontal distance traveled in this time is (16.16*sin 60/9.8)*16.16*0.5 = 11.53.
The boy is standing 11.53 m from the house.