1. A Ball is thrown upward at an angle 90 degree to the ground. At the top of th
ID: 2170248 • Letter: 1
Question
1. A Ball is thrown upward at an angle 90 degree to the ground. At the top of the motion, which quantities are zero?a. acceleration b. displacement c. velocity d. force e. none of these
2. A mass of 6 kg is dropped from a height of 1 meter (the force on it from the earth is 60 Newtons). The earth is
10^24 times more massive. The force of the mass pulling on the earth is approximately how many Newtons?
a. 60 b. 60 x 10^24 c. (1/6) x 10^-24 d. (1/60) x 10^-24 e. none of these
3. A block is resting on a frictionless horizontal surface. There is an upward (
Explanation / Answer
1) Thinking if it as a right angled triangle the vertical component is: 40sin50=30.64 Using the equation: v=u+at (v is final speed (we will calculate the time it takes to get to its highest point in the trajectory and so has 0ms^-1, a is gravity (-9.81), u is initial speed 30.64 and t is time) 0=30.64-9.81xt t=3.12s That is the time it takes to reach its highest point, then it will take that long again to hit the ground so: 3.12x2=6.24 b) Its horizontal component is 40cos50=25.71 As its traveling for a total of 6.24s the total distance traveled is 6.24x25.71=160.4m c)It will hit it at the same angle it was projected at so 50 2) Its downwards component is 40sin30=20 use s=ut+1/2(axt^2) a is positive this time because it is acting in the same direction as the velocity 170=20xt+1/2(9.81xt^2) t=4.19 (it will also give a result of -8.26 however we will only consider the positive result) b)the horizontal component is 40cos30=34.64 As it is traveling for a total time of 4.19s before hitting the ground 34.64x4.19=145.15m c)At an angle of 60 degrees (90-30=60) 3)horizontal component is 20cos40=15.32 Calculating the time it takes to travel 8m and so hit the wall 8/15.32=0.522s The vertical component is 20sin40=12.86 Working out how high the water is traveled after 0.522s will give you how high it strikes: 0.522x12.86=6.71m 4) 12.02ms^-1 This one is quite tricky but I think i got the correct answer: Ok so we will call the initial speed X The horizontal speed it Xcos30 The time taken for the ball to travel 20m is 20/(Xcos30) Now we need to consider the vertical speed: using v^2=u^2=2as u is the inital speed, a is gravity and s is displacement so v^2=x^2-98.1 v=(x^2-98.1)^1/2 Now that we have the time and the final speed we can use the equation v=u+at (remember u is inital speed so can be written as x) subsituting v and t we get (x^2-98.1)^1/2=x+-9.81x20/(xcos30) squaring both sides and doing some rearranging we get the equation 355x^2-51325.92=0 solving the equation we get x=12.02
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.