Answer to Question #271483 in Mechanics | Relativity for JINGJING

Question

SA1. A pebble is projected vertically upward using a sling shot at a velocity of 50.0 m/s. What is its velocity at a height of 2.0 m? 

SA2. At what height of the flight of the pebble in SA1 shall it have a velocity of 12.0 m/s? 

SA3. How long shall the pebble in SA1 attain a velocity of 0.0 m/s? 

SA4. Where shall the pebble in SA1 be located 4.0 seconds after it was launched? 

SA5. How far shall the pebble in SA1 be from its point of origin at the time described in SA4?

 SA6. A ball is projected vertically upward at a velocity of 15.0 m/s. At what position shall it be located at a velocity of – 10.0 m/s?

Answer 1

1.

$v=v_0-gt$

$h=v_0t-gt^2/2=2$ m

$50t-4.9t^2-2=0$

$t=\frac{50\pm\sqrt{2500-400}}{9.8}$

$t_1=0.4$ s, $t_2=9.8$ s

$v_1=50-9.8\cdot0.4=46.1$ m/s, going up

$v_2=gt-v_0=9.8\cdot9.8-50=46.1$ m/s, going down

2.

$t=(v_0-v)/g=(50-12)/9.8=3.9$ s

$h=50\cdot3.9-4.9\cdot3.9^2=120.5$ m

3.

$0=50-4.9t$

$t=50/4.9=10.2$ s

4.

$h=50\cdot4-4.9\cdot4^2=121.6$ m

5.

$h=50\cdot4-4.9\cdot4^2=121.6$ m

6.

$t=(v_0-v)/g=(15-10)/9.8=0.5$ s

$h=15\cdot0.5-4.9\cdot0.5^2=6.3$ m