Question

a projectile is launched with a speed of 140 m/s at an angle of 45 degrees above the horizontal, from a height of 1500 above the ground.how far does it travel in the horizontal direction before hitting the ground.

Answer 1

there are different ways of solving this problem.
the way would do:
equation of motion on y direction:
\[y=v_{0y}t-gt^2/2\]
giving y=-1500m, you can find time when the object hits the ground, and then use
\[x=v_{0x}t\]
to find displacement in horizontal direction:
\[-1500=140*\sqrt{2}/2t-9.81/2t^2\]
move things around:
\[4.9t^2-99t-1500=0\]
this is second order equation in the form ax^2+bx+c=0
solution:
\[t_{1,2}=(-b\pm\sqrt{b^2-4ac})/2a\]
you should get minus for one of t's, that is not possible. Thus use other one, and put in
\[x=v_xt\]

Answer 2

what about the other approach?

Answer 3

can we make use parabola equation since the graph of projectile is simply a parabola?

Answer 4

what do you mean by using parabola equation?

Answer 5

starting from back of posted solution, at the end you need to solve
\[x=v_x*t\]
here you can find t by different ways.
second way would be: you can think this motion in 2 parts, first from starting point to maximum height it would go
\[t_{tomax}=v_{0y}/g\]
then from top it does free fall to ground. distance it goes up to max:
\[h_{tomax}=v_{0y}^2/2g\]
time of free fall from max to ground
\[t_{from max to ground}=\sqrt{2(h_{tomax}+1500)/g}\]
then
\[x=v_{0x}(t_{tommax}+t_{from max to ground})\]

Answer 6

sry what is √2/2t in the equation?

Answer 7

are you asking:
\[t=\sqrt{2h/g}\]
it is from free fall equation:
\[h=gt^2/2\]

Answer 8

i mean in −1500=140∗√2/2t−9.81/2t^2,
i know 140 is V0y but what about √2/2t?

Answer 9

o you need
\[v_0y=v_0\sin{\theta}\]

Answer 10

\[\sin{45}=\sqrt{2}/2\]

Answer 11

OH!!!! I C I C thx!!!