Yes! thank you both! this really helps1
For this problem: A person throws a ball upward with an initial velocity of 15 m per second. In order to calculate how high the ball goes, drwls gave me this formula below: It travels upwards until the vertical velocity component becomes zero. The calculaitons are below: (1/2)...
How would you reword the terms of this equation: (1/2)M Vo^2 = M g H Where H is the maximum height above the thrower's hand. H = Vo²/(2 g) = 11.5 m Could you substitute the letters H & M with Y, Yo, V, Vo, G, T, or use any of these four newton's equtions: V = Vo +...
thanx =) for all your help!
so, then to calculate the distance, the formula would be 50 m =(1/2)(9.8m/s^2)(t^2) and you would neglect the negative signs, but then instead of dividing 50m by -90.8m/s^2, how would u use that formula to find the value of t? Sorry, i am just really confused regarding all of ...
-Suppose that a person is standing on the edge of a cliff 50 m high, so that the ball in his/her hand can fall to the base of the cliff. Questions (a)-How long does it take to reach the base of the cliff? (b)- What is the total distance traveled by the ball? Attempted Answers ...
1 kg is equivilant to about 2.2 lbs. So if you would give 0.1 g per kg, you would convert this to lbs which would give you approx. 0.045 gram/lb. Now simply multiply. 0.045 x 50 = 2.25 grams. =)
thanx so much! im finally starting to get this!
-A person throws a ball upward into the air with an initial velocity of 15 m per second. - How could you calculate how high the ball goes and how long the ball is in the air before it comes to the thrower's hand?
For Further Reading