Question

A boy throws a ball with a velocity of v_^=(3i^+4j^)m/s towards a wall which is 1.2 m from boy where it hits at height h1 . if the launch velocity is v_^= (5i^ +4j^ ) m/s and h2 is the corresponding height then (consider height of the wall >h1 and h2
1)h2=h1
2)h2>h1
3)h2<h1
4) Can't determine

plz anwer this question , fast...​

Answer 1

Answer:

3) h2<h1

Explanation:

FOR BALL 1:

Horizontal distance travelled by ball 1 = 1.2m

3m/s is its velocity in x direction, hence 3$t_{1}$=1.2

hence, $t_{1}$=0.4 s

• $h_{1}$=$u_{y}t_{1} -\frac{1}{2}gt_{1} ^{2}$  =  4(0.4)-5(0.16) = 1.6-0.8 = 0.8m

For Ball 2

Horizontal distance travelled by ball 2 = 1.2

5m/s is its velocity in x direction, hence 5$t_{2}$=1.2

hence, $t_{2}$=0.24 s

• $h_{2}$ = $u_{y}t_{2} -\frac{1}{2}gt_{2} ^{2}$  =4(0.24) -5(0.0576) =0.96-0.288=0.672m

After comparing both the heights

• $h_{2}$ < $h_{1}$

Answer 2

Explanation:

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