Question

A stone is thrown up from a platform with an initial velocity of 19.8 m/s. if

the top of the platform is taken to be the ground level, calculate, (a) the time
taken to reach the maximum height, (b) the maximum height reach by the
stone (c) its velocity just before reaching the ground and (d) total time taken
to reach back to the top of the platform

Answer 1

The physics problem regarding a stone thrown from a platform involves calculating the time to reach maximum height, the maximum height itself, the velocity just before hitting the ground, and the total time to return to the initial level, using the kinematic equations for motion under uniform acceleration.

Step-by-step explanation:

When a stone is thrown up from a platform with an initial velocity, we can use the kinematic equations for motion under uniform acceleration to solve for the quantities in question. Assuming the upward direction is positive and acceleration due to gravity (g) is 9.8 m/s2 downward, we can start calculating:

1. (a) The time taken to reach the maximum height can be found using the equation v = u + at, where v is the final velocity (0 m/s at the maximum height), u is the initial velocity (19.8 m/s), and a is the acceleration (-9.8 m/s2 due to gravity). Solving for t gives t = (v - u)/a.
2. (b) The maximum height reached by the stone can be calculated using the equation s = ut + 0.5at2, where s is the displacement (maximum height), and t is the time calculated in part (a).
3. (c) The velocity just before reaching the ground is the same as the initial velocity but in the opposite direction because the motion is symmetrical. Thus, it is -19.8 m/s.
4. (d) The total time taken to reach back to the top of the platform is double the time taken to reach the maximum height, as the ascent and descent times are equal.

Answer 2

a) 2.02 s

b) 20.0 m

c) -19.8 m/s

d) 4.04 s

Step-by-step explanation:

Given:

v₀ = 19.8 m/s

a = -9.8 m/s²

a) Find t when v = 0 m/s.

v = at + v₀

(0 m/s) = (-9.8 m/s²) t + (19.8 m/s)

t = 2.02 s

b) Find Δy when v = 0 m/s.

v² = v₀² + 2aΔy

(0 m/s)² = (19.8 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 20.0 m

c) Find v when Δy = 0 m.

v² = v₀² + 2aΔy

v² = (19.8 m/s)² + 2 (-9.8 m/s²) (0 m)

v = -19.8 m/s

d) Find t when Δy = 0 m.

Δy = v₀ t + ½ at²

(0 m) = (19.8 m/s) t + ½ (-9.8 m/s²) t²

t = 4.04 s