A ball is thrown at 20 m/s at an angle of 30 degrees up above the horizontal. Th

Question

A ball is thrown at 20 m/s at an angle of 30 degrees up above the
horizontal. The only thing acting on the ball is gravity, so the ball has an
acceleration of 9.8 m/s2 in the downwards (-y) direction. What is the velocity of the
ball two seconds later? (Assume that it is still in the air at 2 seconds.)
First identify all relevant horizontal and vertical quantities (given/known/implied
or asked for), and those quantities that need to be decomposed into x- and ycomponents.
Horizontal quantities: ________________________________________
Vertical quantities: __________________________________________
Quantities to decompose into components: ________________________
Next, find the vector components, and finally solve the problem (as two separate
1D problems). You can leave your answer in component form.

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 Hope this helps :)

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