MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Shane Macguire Grade: upto college level
         A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection (a) as seen from the truck, (b) as seen from the road.
3 years ago

Answers : (1)

Deepak Patra
askIITians Faculty
474 Points
										Sol. . a) As seen from the truck the ball moves vertically upward comes back. Time taken = time 

taken by truck to cover 58.8 m.
∴ time = s/v = 58.8/14.7 = 4 sec. (V = 14.7 m/s of truck)
u = ?, v = 0, g = –9.8 m/s2 (going upward), t = 4/2 = 2 sec.
v = u + at ⇒ 0 = u – 9.8 × 2 ⇒ u = 19.6 m/s. (vertical upward velocity).
b) From road it seems to be projectile motion.
Total time of flight = 4 sec
In this time horizontal range covered 58.8 m = x
∴ X = u cos θ t
⇒ u cos θ = 14.7 …(1)
Taking vertical component of velocity into consideration.

y = (0^2- (19.6)^2)/(2 x (-9.8)) = 19.6 m [from (a)]
∴ y = u sin θ t – 1/2 gt2
⇒ 19.6 = u sin θ (2) – 1/2 (9.8)22 ⇒ 2u sin θ = 19.6 × 2
⇒ u sin θ = 19.6 …(ii)
(u sin⁡θ)/(u cos⁡θ ) = tan θ 19.6/14.7 = 1.333
⇒ θ = tan–1 (1.333) = 53°
Again u cos θ = 14.7
⇒ u = 14.7/(u cos⁡〖53°〗 ) = 24.42 m/s.
The speed of ball is 42.42 m/s at an angle 53° with horizontal as seen from the road.
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details