Question

1)A clay plate at a shooting range is launched straight up into the air from a device 1.10 m above the ground. After 1.40 s the clay plate is pulverized at a height of 15.0 m. What initial speed was the clay plate launched upward?
2)A person is walking along an icy road. They are able to keep a good walking pace of 1.40 m/s. The road suddenly slopes downward and the person slips down the 38.0 m slope. It takes them 7.10 s to slide all the way down the slope. What is the angle of the slope?

Please, if you could help me out with these and also provide explanations, that would be greatly appreciated!

Answer 1

#1

Initial position of clay plate is y1 = 1.10 m

final position of clay plate is y2 = 15 m

time taken t = 1.40 s

acceleration of plate a = -9.8 m/s^2

now we can use kinematics here to find the initial speed

$y_2 - y_1 = v_i*t + \frac{1}{2}at^2$

$15 - 1.1 = v_i* 1.40 - \frac{1}{2}*9.8*1.40^2$

$v_i = 16.8 m/s$

so initial speed of launch is 16.8 m/s

#2

initial speed is given as v1 = 1.40 m/s

distance moved by the person = 38 m

time taken = 7.10 s

now the acceleration can be found by kinematics

$d = v_i* t + \frac{1}{2}at^2$

$38 = 1.40 * 7.10 + \frac{1}{2}a*7.1^2$

$38 = 9.94 + 25.2 *a$

$a = 1.11$

so here on inclined plane we know that

$a = g sin\theta$

$1.11 = 9.8 sin\theta$

$\theta = 6.5 degree$

so the angle of inclined plane is 6.5 degree