A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 7.90 m/s at an angle of 17.0° below the horizontal. It strikes the ground 4.00 s later.
Find the height from which the ball was thrown.
I cannot get the correct answer using the standard constant accel equations. Not sure what I am doing wrong.
2007-02-13 07:12:38 · 5 answers · asked by Matt F 1 in Science & Mathematics ➔ Physics
v=vo+g*t
v(horizontal) = 7.90*cos(17.0)
vi(vertical) = 7.90*sin(17.0)
in the vertical direction we can figure out how fast it is going when it hits the ground.
vf(v) = vi(v)+g*t
taking the final velocity we can use the distance equation to find the initial hieght, h@ the ground being 0.
h=vf(v)*t+1/2gt^2
2007-02-13 07:17:29 · answer #1 · answered by Anonymous · 0⤊ 0⤋
To do physics issues, basically write down what you comprehend. See what equations you are able to plug issues into to remedy for yet another variable, then plug each and all of the variables into yet another equation or 2 and that would desire to offer you your very final answer. a million. the area is the comparable going up and down the hill. d = rt supplies us d = 37*t_1 = sixty six*t_2 fixing for the cases supplies t_1 = d/37 and t_2 = d/sixty six to locate the known velocity, you basically divide finished distance via finished time. subsequently the finished distance is 2nd, and the time is t_1 + t_2. s_ave = 2nd/(t_1 + t_2) = 2nd/(d/37 + d/sixty six) and the d's cancel, so basically simplify. 2. v_0 = 3 m/s a = 2.5 m/s^2 x_0 = 0 x = 8 m Plug into v^2 = (v_0)^2 + 2a(x - x_0) 3. v_0 = 8 m/s t = 6 s x0 = 0 x = 60 m Plug into x = x_0 + v_0 * t + 0.5 * a * t^2 4. Use v^2 = (v_0)^2 + 2a(x - x_0) the place v = 0 via fact she has come to a provide up 5. Is the particle accelerating a million.5 m/s^2? Typo? x_ 0 = 0 x = 22 m a = a million.5 m/s^2 t = 4 s Plug into x = x_0 + v_0 * t + 0.5 * a * t^2 and remedy for v_0 Then plug into v^2 = (v_0)^2 + 2a(x - x_0)
2016-10-02 02:14:40 · answer #2 · answered by Anonymous · 0⤊ 0⤋
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2007-02-13 07:18:42 · answer #3 · answered by god knows and sees else Yahoo 6 · 0⤊ 0⤋
assume upward is +, downward is -
we only care about the vertical component since that's all height is part of:
velocity of downward component = -7.90m/s(sin(17))
accel of downward component = -9.8m/s^2
so
distance = 0 -7.90m/s(sin(17)) -1/2(-9.8m/s^2)
distance (in positive) is the height
see if you get this right.
2007-02-13 07:17:43 · answer #4 · answered by yungr01 3 · 0⤊ 0⤋
h=gt^2/2+tvsin17*
t=4
2007-02-13 07:23:53 · answer #5 · answered by a 3 · 0⤊ 0⤋