Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
a baseball is hit at a height 1m and then caught at the same height. It travels alongside a wall, moving up past the top of the wall 1.00 s after it is hit and then down past the top of the wall 4.00 s later, at distance 50m farther along the wall. Please explain.
(a) What horizontal distance is traveled by the ball from hit to catch?
(b) magnitude
(c) angle (relative to the horizontal) of the ball's velocity just after being hit?
(d) How high is the wall?
Let us assume that the ball was hit with velocity V making an angle 'b' with the horizontal.
So it horizontal component of velocity will be constant and will be equal to v cos b.
Now we know the ball took one sec to move pas the wall and then three more seconds to move the distance when it was above the wall. So it must have taken one more second till it was caught; making the total time interval to 5 seconds.
Also we know the ball travelled 50m in three seconds with a constant horizontal velocity; there fore the horizontal component must be equal to 50/3 m/sec.
Therefor total horizontal distance will be equal to 50/3*5= 250/3 m.
Now if horizontal component is 50/3=v cos b then v sin b= say x.. this will give you the height of the wall as this velocity took one sec to cover the height of the wall.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !