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tini12
 one year ago
Somewhere in the vast flat tundra of planet Tehar, a projectile is launched from the ground at an angle of 20 degrees. It reaches the maximum height of 15 m. The acceleration due to gravity is 15 m/s^2. Find the minimum speed of the projectile during the flight. Enter the answer in "m/s".
tini12
 one year ago
Somewhere in the vast flat tundra of planet Tehar, a projectile is launched from the ground at an angle of 20 degrees. It reaches the maximum height of 15 m. The acceleration due to gravity is 15 m/s^2. Find the minimum speed of the projectile during the flight. Enter the answer in "m/s".

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tini12
 one year ago
Best ResponseYou've already chosen the best response.0You are participating in an unusual drag race. You start from rest and try to cover a straight quartermile (400 meters, for the purpose of this problem) in the shortest time possible. Your car is very well built so it can achieve speeds of up to 360 km/h. It is equipped with a very powerful engine and an excellent antilock braking system. The coefficient of static friction between the wheels and the road is about 1.2 (the dragrace tires are made of very soft rubber which is preheated before the start by spinning the wheels; as a result, the wheels achieve the effective coefficient of friction greater than 1). However, there is a crucial difference: you are required, for publicity purposes, to come to a dead stop at the finish line (in a typical drag race, you would fly right through that finish line!). speed up at the maximum possible rate for the first half of the distance; then slow down at the same rate What is the shortest possible time needed to complete the race (in seconds)?

Festinger
 one year ago
Best ResponseYou've already chosen the best response.0For the first problem the flat tundra thing. Recall projectile motion. At it's peak, there is horizontal motion. At any other point, there is vertical velocity. Since there is no horizontal acceleration, and by bringing back newton's first law, the minimum speed is where the vertical speed is zero, and thus finding the horizontal speed would suffice. Now the problem to find the horizontal distance. Recall that: \[v^2=u^2+2as\] So we find that u=21.2m/s Resolving, \[V_{vertical}=V_{0}Sin 20^{o}=21.2m/s\]and \[V_{horizontal}=V_{minimum}=V_{0}Cos20^{o}\] Solving for , v minimum is then 58.3m/s
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