Given:
Dy= 20 m
Vi = 5.0 m/s horizontally
A=9.81 m/s^2

Find:
Horizontal displacement

Solution:
D=ViT+(1/2)AT^2
Dy=(1/2)AT^2
T^2=Dy/(1/2)A
T=sqrt(Dy/(1/2)A)
T=sqrt(20/4.905)
T=2.0s

Dx=ViT
Dx=(5.0)(2.0)
Dx=10. meters

7 0

3 years ago

Can some answer 7 a and b please

pav-90 [236]

Answer:

a.After $15$ second Mr Comer's speed $=$ $1.7 m/s$

b.Distance travelled by Mr.Comer in $15$ seconds $=$ $18.75\ m$

Explanation:

a. Lets recall our first equation of motion $V_{f} =V_{i} + at$

Now we know that $V_{i} = 0.8m/s$ , $a = 0.06\ m/s^2$ and

$t = 15 \ sec$

Plugging the values we have.

$V_{f} =V_{i} + at$

$V_{f} =0.8 + 0.06 \times 15$

$V_{f} =0.8+ 0.9$

$V_{f} =1.7 m/s$

Then Mr.Comer's speed after $15$ sec $=$ $1.7ms^-1$

Lets find the distance and recall our third equation of motion.

$V_{f} ^2-V{i}^2 = 2as$

So $s =$ distance covered.

Dividing both sides with 2a we have.

$\frac{V_{f} ^2-V{i}^2}{2a} = 2as$

Plugging the values.

$\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as$

$s= 18.75\ m$

So Mr.Comer will travel a distance of $s= 18.75\ m$ .

4 0

3 years ago

What is the kinetic energy f a 25kg object movingat a velocity of 10m/s?

alex41 [277]

Using K.E=1/2MV^2
answer is 125joules

5 0

4 years ago