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anonymous
 one year ago
A girl throws a waterfilled ballon at an angle of 50 degrees above the horizontal with a speed of 15 m/s. The horizontal component of the balloon's velocity is directed toward a car that is approaching the girl at a constant speed of 10 m/s. If the balloon is to hit the car at the same height at which it leaves her hand, what is the maximum distance the car can be from the girl when the balloon is thrown? Ignore air resistance.
anonymous
 one year ago
A girl throws a waterfilled ballon at an angle of 50 degrees above the horizontal with a speed of 15 m/s. The horizontal component of the balloon's velocity is directed toward a car that is approaching the girl at a constant speed of 10 m/s. If the balloon is to hit the car at the same height at which it leaves her hand, what is the maximum distance the car can be from the girl when the balloon is thrown? Ignore air resistance.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Jack1 If you want to help. Feel free or we can do it tomorrow! It is already midnight in my timezone.

rajat97
 one year ago
Best ResponseYou've already chosen the best response.1that's a good question here, you need to break the projectile motion as usual into two directionsthe vertical direction and the horizontal direction uniformly accelerated motion in the vertical direction with acceleration g and uniform motion in horizontal direction now we need the ballon to be thrown such that it hits the car and it is said that 'balloon is to hit the car at the same height at which it leaves her hand' so the ballon will complete the range of the projectile motion it traces and hits the car, so the time of the projectile motion should be equal to the time taken by the car to go from it's initial position to the point where the ballon completes it's range so, time of projectile motion (t) = distance of car from the end point of the projectile motion(x)/10m/s we know that the formula for the time of projectile motion = 2usin(theta)/g where, u is the initial speed of the projectile, theta is the angle of projection from the horizontal and g is the acceleration due to gravity so we get t= 2*15*sin(50deg)/9.8 = 2.34s thus the distance travelled by the car is speed*time = 10*2.34 = 23.4m and the car's distance from the point of projection i.e. from the girl = Range of projectile  distance travelled by the car now, we need to find the range we know that the fromula for range of a projectile = R = u^(2) * sin(2theta)/g where, all the alphabets and symbols have their usual meaning so we get the range = R = 22.61m so the distance of the car from the girl = 22.61  23.4 = 0.79m the '' sign indicates that the car starts off from behind the girl this may not be a very clear explanation so if you have any problem, feel free to ask! hope this helps you:)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hey I just got online! Will read and understand your text. Thank you and God bless mate.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439698283742:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is my drawing right?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439723801163:dw

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0the car starts from A and the collision takes place at B for the car v = d/t so 10 = AB / t for the projectile 15cos 50 = BC / t the time involved is the same for the car and the projectile so AB / 10 = BC / 15cos50 AB/10 = BC / 9.642

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0to find the time we can use the vertical velocity of the ballonon v = u + at 0 = 15sin50  9.81t t = 15sin50/9.81 = 1.1713 seconds  this is for half the distance so time from girl to car = 2 * 1.1713 = 2.3426 seconds using the AB/10 = t AB = 10*2.3426 = 23.43 m and BC = 9.642*2.3426 = 19.60 m so total distance = 43.03 m
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