Question

16. A ball is thrown vertically upwards and reaches a maximum height in 3 sec Calcula
a. The velocity with which it was thrown upwards.
b. Maximum height attained by the ball.​

Answer 1

Answer:

a is 30 and b is 135

Explanation:

t=u/g that is 3sec=x/g g= 10 so x =30 s=ut+1/2at*2 u=30 t=3 a=10 so s=30×3+1/2×10 3*2 =135

Answer 2

Answer:

ball is thrown upward and reaches a maximum height in 3second

so the velocity with which the ball was thrown

$\</strong><strong>g</strong><strong>r</strong><strong>e</strong><strong>e</strong><strong>n</strong><strong>{</strong><strong>v</strong><strong>=</strong><strong>u</strong><strong>+</strong><strong>at</strong><strong>}$

v=final velocity=0

u=initial velocity

accelaration(a)=9.8m/s^2

time taken(t)=3s

$0 = u + (3)(9.8) \\ - 29.4mpersecond = u$

the distance convered by the ball is

$d = ut + \frac{1}{2} at {}^{2}$

$d = ( - 29.4)(3) + \frac{1}{2} ( - 29.4)( 3) {}^{2} \\ d = - 88.3 - 132.3 \\ d = - 220.6m$