1.
Answer:
y = 11.48 m
x = 13.0 m
Explanation:
Components of initial velocity is given as
[tex]v_x = 7.2 cos25 = 6.52 m/s[/tex]
[tex]v_y = 7.2 sin25 = 3.04 m/s[/tex]
Now after t = 2 s the vertical position is given as
[tex]y = y_0 + v_y t + \frac{1}{2}a_y t^2[/tex]
[tex]y = 25 + 3.04(2) - \frac{1}{2}(9.8)(2^2)[/tex]
[tex]y = 11.48 m[/tex]
Now horizontal position is given as
[tex]x = v_x t[/tex]
[tex]x = 6.52 \times 2[/tex]
[tex]x = 13.04 m[/tex]
2.
Answer:
d = 26.6 m
Explanation:
Initial position on y axis is given as
[tex]y = 1.5 m[/tex]
velocity of ball in y direction
[tex]v_y = 0[/tex]
now we have
[tex]\Delta y = v_y t + \frac{1}{2}gt^2[/tex]
[tex]1.5 = \frac{1}{2}(9.8) t^2[/tex]
[tex]t = 0.55 s[/tex]
now the distance moved by the ball in horizontal direction is given as
[tex]d = v_x t[/tex]
[tex]d = 48.1 \times 0.55[/tex]
[tex]d = 26.6 m[/tex]
3
Answer:
[tex]A + B = 12.42\hat i + 0.35 \hat j[/tex]
Explanation:
Here we can see the two vectors inclined at different angles
so two components of vector A is given as
[tex]A = 11.3 cos21\hat i - 11.3 sin21\hat j[/tex]
[tex]A = 10.55 \hat i - 4.05 \hat j[/tex]
Similarly for other vector B we have
[tex]B = 4.78cos67 \hat i + 4.78 sin67\hat j[/tex]
[tex]B = 1.87\hat i + 4.4 \hat j[/tex]
now we need to find A + B
so we have
[tex]A + B = (10.55\hat i - 4.05\hat j) + (1.87\hat i + 4.4 \hat j)[/tex]
[tex]A + B = 12.42\hat i + 0.35 \hat j[/tex]
4.
Answer:
d = 81.86 m
Explanation:
Displacement of airplane is given as
[tex]d_1[/tex] = 72 km in direction 30 degree East of North
[tex]d_1 = 72sin30\hat i + 72cos30\hat j[/tex]
[tex]d_1 = 36\hat i + 62.35\hat j[/tex]
[tex]d_2[/tex] = 48 km South
[tex]d_2 = -48\hat j[/tex]
[tex]d_3[/tex] = 100 km in direction 30 degree North of West
[tex]d_3 = -100 cos30\hat i + 100 sin30\hat j[/tex]
[tex]d_3 = -86.6\hat i + 50\hat j[/tex]
so net displacement is given as
[tex]d = d_1 + d_2 + d_3[/tex]
[tex]d = 36\hat i + 62.35\hat j - 48\hat j - 86.6\hat i + 50 \hat j[/tex]
[tex]d = -50.6\hat i + 64.35\hat j[/tex]
now magnitude of displacement is given as
[tex]d = \sqrt{50.6^2 + 64.35^2}[/tex]
[tex]d = 81.86 m[/tex]
5.
Answer:
1) final speed at which it will hit the ground again
2) acceleration during the motion of the ball
Explanation:
As we know that the speed at which the ball is thrown is always same to the speed by which it will hit back on the ground
so we know that final speed will be same as initial speed
Also we know that during the motion the acceleration of ball is due to gravity so it will be
[tex]a = - g[/tex]
