Question

A stone thrown up is caught by the thrower after 6s. How

high did it go and what was its position 4s after start?

(g=9.8m/s2).​

Answer 1

Answer:

it went 45 m  high above the ground and it's position 4s after start = 40 m

Answer 2

Position of ball from ground is 39.2m.

What is g in projectile motion?

• In projectile motion, we try to take account motion in both the horizontal and vertical directions.
• We can come up with the basic formulas for simple projectile motion by assuming that the acceleration in the x direction is zero, and the acceleration in the y direction is -g where g is the acceleration due to gravity.

What is the formula of maximum height of projectile?

• The maximum height h reached by the projectile is equal to one-half of H, the altitude of this triangle. = H – ½H so h = H/2, which is the desired result.

According to the question:

Total time of flight = 6s.

According due to gravity = 9.8m/s²

For a body thrown vertically up with initial velocity u,

Total time of flight

$\mathrm{T}=\frac{2 \mathrm{u}}{\mathrm{g}}$

Maximum height reached

$\mathrm{H}=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}$

Maximum helght:

$\begin{aligned}&\mathrm{H}=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}} \\&\mathrm{H}=\frac{\mathrm{g}^{2} \mathrm{~T}^{2}}{4 \times 2 \mathrm{~g}} \\&\mathrm{H}=\frac{\mathrm{gT}^{2}}{8} \\&\mathrm{H}=\frac{9.8 \tilde{A} 36}{8} \\&\mathrm{H}=44.1 \mathrm{~m}\end{aligned}$

Position of ball at t=4 s:

Time of ascent $=\frac{\text { Total time of flight }}{2}$

Time of ascent $=\frac{6}{2}=3 \mathrm{~s}$

Ball will reach at maximum height at t=$3 \mathrm{~s}$

$\begin{aligned}&d=u t+\frac{1}{2} g t^{2} \\&d=0+\frac{1}{2} g(4-3)^{2} \\&d=0.5 \times 9.8 \times 1 \\&d=4.9 \mathrm{~m}\end{aligned}\\At\ t=4 s :$

Position of ball from ground = 44.1-4.9 = 39.2m

Hence, The distance between the ball and the ground is 39.2 metres.

Learn more about height of projectile here,

https://brainly.in/question/49564502?msp_poc_exp=5

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