13. Given a position vector P=-4i + 6j What is the cosine of the angle between p and (a) the x-axis?. (b) the y-axis?. The cosine of the angle between a vector and the positive coordinate axis is called direction cosine of the vector.
14. A ball is thrown vertically upwards with a velocity of 10m/s from the balcony of a tall building. The balcony is 15m above the ground and gravitational acceleration is 10m/s². Find (a) the time required
for the ball to hit the ground, and (b) the velocity with which it hits the ground.
15. What vector must be added to vector C of 10km, east in order to give a resultant vector of?
A/15km,east.
B/ 15 km, West.
13.
(a) To find the cosine of the angle between the vector P and the x-axis, we need to find the projection of vector P onto the x-axis. Since the x-axis is a horizontal line, the projection of vector P onto the x-axis will be the x-component of vector P.
The x-component of vector P is -4.
Therefore, the cosine of the angle between vector P and the x-axis is given by the formula:
cos(angle) = x-component of P / magnitude of P.
In this case:
cos(angle) = -4 / sqrt((-4)^2 + 6^2) = -4 / sqrt(52) = -4 / 2sqrt(13) = -2sqrt(13) / 13.
So, the cosine of the angle between vector P and the x-axis is -2sqrt(13) / 13.
(b) Similarly, to find the cosine of the angle between the vector P and the y-axis, we need to find the projection of vector P onto the y-axis. Since the y-axis is a vertical line, the projection of vector P onto the y-axis will be the y-component of vector P.
The y-component of vector P is 6.
Therefore, the cosine of the angle between vector P and the y-axis is given by the formula:
cos(angle) = y-component of P / magnitude of P.
In this case:
cos(angle) = 6 / sqrt((-4)^2 + 6^2) = 6 / sqrt(52) = 6 / 2sqrt(13) = 3sqrt(13) / 13.
So, the cosine of the angle between vector P and the y-axis is 3sqrt(13) / 13.
14.
(a) The equation for the height of an object thrown vertically upwards can be given as:
y = ut - (1/2)gt^2
where y is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
In this case, y = 15m (since the balcony is 15m above the ground), u = 10m/s, and g = -10m/s^2 (since the acceleration due to gravity is negative as it acts downwards).
Setting y = 0 (since the ball hits the ground), we can solve for t:
0 = 10t - (1/2)(-10)t^2
Multiplying through by 2:
0 = 20t + 10t^2
Rearranging the equation:
10t^2 + 20t = 0
Factoring out a common factor of 10t:
10t(t + 2) = 0
Setting each factor equal to zero:
10t = 0 or t + 2 = 0
Solving for t:
t = 0 or t = -2
Since time cannot be negative, the ball hits the ground at t = 0.
Therefore, the time required for the ball to hit the ground is 0 seconds.
(b) The velocity with which the ball hits the ground is equal to the final velocity when it reaches the ground.
Using the equation for velocity:
v = u + gt
where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
In this case, u = 10m/s, g = -10m/s^2, and t = 0 (since the ball hits the ground at t = 0).
Plugging in the values:
v = 10m/s + (-10m/s^2)(0) = 10m/s
Therefore, the velocity with which the ball hits the ground is 10m/s.
15.
To find the vector that must be added to vector C of 10km, east in order to give a resultant vector of A/15km, east or B/15km, west, we need to subtract vector C from vector A or B, respectively.
(a) To find the vector that must be added to vector C to give a resultant vector of 15km, east, we subtract vector C from vector A:
15km, east - 10km, east = 5km, east.
Therefore, the vector that must be added to vector C of 10km, east to give a resultant vector of 15km, east is 5km, east.
(b) To find the vector that must be added to vector C to give a resultant vector of 15km, west, we subtract vector C from vector B:
15km, west - 10km, east = 5km, west.
Therefore, the vector that must be added to vector C of 10km, east to give a resultant vector of 15km, west is 5km, west.