Question

An object is thrown upwards with a speed of 16m/s. How long does it take it to reach a height of 7m

Answer 1

Given:-

• Initial velocity ,u = 16m/s

• Final velocity ,v = 0m/s

• Acceleration due to gravity ,a = 9.8m/s²

• Height ,h = 7m

To Find:-

• Time taken ,t

Solution:-

We can find the time taken by using 2nd Equation of motion as well as 1st Equation of motion. So we use here 1st Equation of motion .

• v = u + at

Substitute the value we get

➨ 0 = 16 +(- 9.8)×t

➨ -16 = -9.8×t

➨ t = -16/-9.8

➨ t = 16/9.8

➨ t = 1.63 s

Therefore, the time to reach the height of 7m is 1.63 second.

Additional Information!!

Some equation of motion are

• v = u +at

• s = ut +1/2at².

• v² = u² + 2as

Here,

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement or Distance

t is the time taken

Answer 2

Answer:

$\huge \bf \bigstar \: solution$

According to 1 st equation of motion

$\huge \bf \: v \: = u + at$

Here,

$\sf \implies \: initial \: velocity(u) \: = 16 \: m \: per \: s$

$\sf \implies \: final \: velocity \: (v) = 0 \: m \: per \: s$

$\sf \implies \: acceleration \: = 9.8m²$

$\sf \implies \: height \: = 7 \: m$

Now,

According to the formula

$\sf 0 = 16 + {9.8m \: } \times t$

$\sf - 16 = { - 9.8}t$

$\sf t \: = \frac{ - 16}{ - 9.8}$

$\sf \: t \: = \frac{16}{9.8}$

$\sf \: t \: = 1.63\: second$

$\huge \bf \: time \: taken = 1.63 \: sec$