Question

A body is thrown up with velocity 50 m/s calculate the maximum height reached and time taken to reach the maximum height (a =10m2

Answer 1

Answer:

ghvcjjhhgvghall bhhgghell nap in the morning so I can get it from me for the day before the end of my friends and I don't know what I can do that to me that I have to go to the gym and I don't know what I

Explanation:

यूके रिकॉर्ड दी समथिंग विच वी कैन डू हिंदी इंवेंटेड पीरियड पीरियड bchildishly dock and the Two Strings to the gym and I don't know what I can do it for you to be with him and his family and friends that have a good day at work today and I have no how much you love me anymore so I can get it from me for the day before the end of my friends and I don't know what to do it again and again and again and again and again and again and again and again in my room is the first place in a while back but it will get a hold of you and I don't know what I can do that ghhhjhhhhhjjjjjjjjjjkjjjjjkllllkhbvvzssadgharm infill I have a good day at work today and I have no idea how to make chocolate powder at home and I don't know what I can do that to me that I have to go to the gym and I don't know what I can do that to me that I have to go to the and I don't know what I can do that to me that I have to go to the gym and I don't want to do it again and again and again and again in my room and I don't know what I can do that to me that I have to me that oghggghjjjjnnnnnnnnjjhhdddedsddseeeddefbnjjHfhGDP hbhgGDP Gk question dikaho I have to go to the gym and I don't know what I can do that to me that I have to go to the gym and I don't know what I can y

Answer 2

Given:-

• Initial velocity (u) = 50m/s

To Find:-

• The maximum height reached by the body.
• Time taken to reach the maximum height.

Solution:-

• Initial velocity (u) = 50m/s
• At maximum height velocity (v) = 0m/s
• Acceleration (a) = g = -10m/s²

Let the maximum height be s

According to the third equation of motion:-

$\rm \dashrightarrow {v}^{2} = {u}^{2} + 2as⇢v </p><p>2</p><p> =u </p><p>2</p><p> +2as</p><p></p><p>Substituting the values:-</p><p></p><p>→ (0)² = (50)² + 2 × (-10) × s</p><p></p><p>→ 0 = 2500 + (-20)s</p><p></p><p>→ 0 = 2500 - 20s</p><p></p><p>→ 20s = 2500</p><p></p><p>→ s = [tex]\dfrac{2500}{20}$

20

2500

→ s = 125

∴Maximum height=125m

Now, let us find time:-

By first equation of motion:-

→ v = u + at

→ 0 = 50 + (-10)t

→ 0 = 50 - 10t

→ 10t = 50

→ t = $\dfrac{50}{10}$

10

50

→ t = 5

∴Time taken to reach maximum height =5sec