Free help with homework

Why join Brainly?

  • ask questions about your assignment
  • get answers with explanations
  • find similar questions

2 balls are thrown up. One ball is thrown 2 seconds after the second ball. If the initial velocity is 40m/s.

Find the height at which they will meet



This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Let first ball is thrown at t = 0s.
second ball is thrown at t = 2s
u = 40 m/s
g = -10 m/s²

let they meet after t seconds at a height H. Then both the balls will be at H.
for 1st ball, H = ut+(1/2)gt²
for 2nd ball, H = u(t-2) + (1/2)g(t-2)²

ut+ \frac{1}{2}gt^2 =u(t-2)+ \frac{1}{2}g(t-2)^2 \\ \\ut+ \frac{1}{2}gt^2 =ut-2u+ \frac{1}{2}g(t^2+4-4t) \\ \\ut+ \frac{1}{2}gt^2 =ut-2u+ \frac{1}{2}gt^2+2g-2gt \\ \\-2u+2g-2gt=0\\ \\u-g+gt=0\\ \\40-(-10)-10t=0\\ \\10t=50\\ \\t=  \frac{50}{10}=5\ seconds

H = ut+ \frac{1}{2}gt^2 \\ \\H=40*5- \frac{1}{2}*10*5^2=\boxed{75\ m}
0 0 0
Thanks my answer was coming 60m thanks for answering it
The Brain
  • The Brain
  • Helper
Not sure about the answer?
Learn more with Brainly!
Having trouble with your homework?
Get free help!
  • 80% of questions are answered in under 10 minutes
  • Answers come with explanations, so that you can learn
  • Answer quality is ensured by our experts