Question

022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The acceleration due to gravity is 9.8 m/s 2 . Answer in units of m.
023 (part 2 of 4) 10.0 points How long does it take to reach its highest point? Answer in units of s.
024 (part 3 of 4) 10.0 points How long does it take the ball to hit the ground after it reaches its highest point? Answer in units of s.
025 (part 4 of 4) 10.0 points What is its velocity when it returns to the level from which it started? Answer in units of m/s.
026 10.0 points An object is released from rest on a planet that has no atmosphere. The object falls freely for 4.88 m in the first second. What is the magnitude of the acceleration due to gravity on the planet? Answer in units of m/s 2 .
027 10.0 points A rocket initially at rest accelerates at a rate of 90 m/s 2 for 0.33 min. What is its speed at the end of this time? Answer in units of m/s.
028 (part 1 of 2) 10.0 points Janet jumps off a high diving platform with a horizontal velocity of 2.92 m/s and lands in the water 2.3 s later. How high is the platform? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.
029 (part 2 of 2) 10.0 points How far from the base of the platform does she land? Answer in units of m.
030 (part 1 of 2) 10.0 points An artillery shell is fired at an angle of 80.9 ◦ above the horizontal ground with an initial

Answer 1

1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

$v_f = 24.5 m/s$

Explanation:

Part 1)

initial speed of the ball upwards

$v_i = 24.5 m/s$

so maximum height of the ball is given by

$H = \frac{v_i^2}{2g}$

$H = \frac{24.5^2}{2(9.80)}$

$H = 30.6 m$

Part 2)

As we know that final speed will be zero at maximum height

so we will have

$v_f - v_i = at$

$0 - 24.5 = (-9.8)t$

$t = 2.5 s$

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

$v_f = 24.5 m/s$

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

$y = \frac{1}{2}at^2$

$4.88 = \frac{1}{2}a(1^2)$

$a = 9.76 m/s^2$

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

$a = 90 m/s^2$

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

$v_f = v_i + at$

$v_f = 0 + (90)(0.33)$

$v_f = 29.7 m/s$

4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}(9.81)(2.3^2)$

$y = 25.95 m$

Part 2)

Distance traveled by it in horizontal direction is given as

$d = v_x t$

$d = 2.92 \times 2.3$

$d = 6.72 m$