Question

Тело бросили вертикально вверх с начальной скоростью 10 м с. на какую высоту поднимется тело? каково время подъема тела?

Answer 1

H=V^2/2*g=100/20=5 м/с
t=V/g=10/10=1 c

Answer 2

Calculation of the Maximum Height and Time of Ascent

To calculate the maximum height and time of ascent of an object thrown vertically upwards with an initial velocity of 10 m/s, we can use the equations of motion.

The equation for the height of the object at any given time is given by:

h = h0 + v0t - (1/2)gt^2 Where: - h is the height of the object at time t - h0 is the initial height (which is 0 in this case since the object is thrown vertically upwards) - v0 is the initial velocity (10 m/s in this case) - g is the acceleration due to gravity (approximately 9.8 m/s^2)

To find the maximum height, we need to find the time at which the object reaches its highest point. At this point, the vertical velocity of the object becomes zero.

Using the equation for vertical velocity:

v = v0 - gt Setting v = 0 and solving for t, we get:

t = v0/g Substituting this value of t into the equation for height, we can find the maximum height:

h_max = h0 + v0(t) - (1/2)g(t)^2 Let's calculate the maximum height and time of ascent using these equations.

Calculation:

Given: - Initial velocity (v0) = 10 m/s - Acceleration due to gravity (g) = 9.8 m/s^2

Using the equation for time of ascent:

t = v0/g

Substituting the values:

t = 10 / 9.8

Calculating the value of t:

t ≈ 1.02 seconds Now, using the equation for maximum height:

h_max = h0 + v0(t) - (1/2)g(t)^2

Substituting the values:

h_max = 0 + 10(1.02) - (1/2)(9.8)(1.02)^2

Calculating the value of h_max:

h_max ≈ 5.1 meters

Answer:

The object will reach a maximum height of approximately 5.1 meters and the time of ascent will be approximately 1.02 seconds.