Need help with b, c and d 3. Along jumper leaves the ground at an angle of 45o a
ID: 1549023 • Letter: N
Question
Need help with b, c and d 3. Along jumper leaves the ground at an angle of 45o above the horizontal, landing 8.25 meters away (from the point at which she jumps). a. Determine her "takeoff speed vo [10] b. Now, while hiking, she comes to the left bank of a river. There is no bridge, and the right [15] bank is 10.0 meters away horizontally and 2.5 meters vertically below. If she jumps from the edge of the left bank, again at an angle of 45° above the horizontal, with the speed determined above, how long, or short, of the opposite bank will she land? Present all detail of the required calculations. c. Determine the magnitude AND direction of the velocity of the long jumper at the she reaches the edge of the left bank and/or the river; ie. determine the magnitude and direction of the velocity of the long jumper at the moment her vertical position 2.5 her initial vertical position. Present detail is of the required calculations. late the maximum height of the long jumper (with respect to the coordinate system d. [10] chosen in Exercise 3b. Present all detail of the required calculations.Explanation / Answer
(A) Range is 8.25 m.
R = v^2 sin(2 theta) / g
8.25 = v^2 sin(90) / 9.8
v = 9 m/s
(B) In vertical,
v0y = 9 sin45 = 6.36 m/s
ay = - 9.8 m/s^2
y = - 2.5 m
y = v0y t + ay t^2 /2
- 2.5 = 6.36t - 4.9 t^2
t = 1.614 sec
In horizontal,
d = (9 cos45) (1.614) = 10.27 m
so she will be 0.27 m long .....ANs
(C) vx = 9 cos45 = 6.36 m/s
vy = 6.36 - (9.8 x 1.614 ) = - 9.45 m/s
magnitude = sqrt(vx^2 + vy^2) = 11.4 m/s
theta = tan^-1(9.45 / 6.36) = 56 deg below the horizontal.
(D) at Hmax, vy = 0
vy^2 - v0y^2 = 2 a d
0^2 - 6.36^2 = 2 (-9.8)(d)
d =2.06 m